id stringlengths 17 53 | domain stringclasses 5
values | category stringclasses 24
values | difficulty stringclasses 3
values | language stringclasses 10
values | problem stringlengths 29 149 | thinking_1 stringlengths 236 493 | solution_1 stringlengths 166 541 | thinking_2 stringlengths 302 467 | solution_2 stringlengths 184 565 | thinking_3 stringlengths 302 463 | solution_3 stringlengths 169 1.14k | metadata dict |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
product_except_self_python_3_678ad7db | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_3_2cf9c5aa | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_3_f97ef2ae | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_3_58cf9657 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_3_ac6d5938 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_3_8f3dba3f | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_3_e6b9151f | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_3_a6f40969 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_3_ad6dd45e | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_3_7c20c93f | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Consider memory constraints for streaming input. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_4_bf280a15 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_4_0e6786ad | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_4_f718e315 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_4_fcabc0b1 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_4_477da841 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_4_6457f001 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_4_af3cbf1f | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_4_0c5505e6 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_4_bd986a5a | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_4_7e645565 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Support both ascending and descending order. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_5_9d6af017 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_5_93261c3c | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_5_cdaa25c3 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_5_51803f91 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_5_95ad2670 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_5_895d2c28 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_5_dce0aa31 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_5_7def4d9d | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_5_fb6f6494 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_5_4b03eabf | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Ensure thread-safety for concurrent access. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_6_f0810d3c | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_6_d1443b18 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_6_a1322bde | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_6_2cab994f | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_6_cb6f21be | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_6_84093071 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_6_7cbded41 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_6_b188919a | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_6_3273defd | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_6_774d8ed5 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Add comprehensive input validation. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_7_b2e91786 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_7_104835ea | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_7_6aad34ac | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_7_33be2cad | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_7_17f50ea3 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_7_d2428839 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_7_0190e71a | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_7_7832b851 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_7_9acb82e7 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_7_ca5a60cb | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Optimize for cache efficiency. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_8_b1df59c5 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_8_d4d6617c | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_8_ac708e6f | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_8_195bcd2c | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_8_e3e967da | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_8_0ca5ce91 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_8_8efd46d8 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_8_53f431c7 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_8_2b07546b | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_8_a858449e | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Use zero-based indexing throughout. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_9_302917ee | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_9_d22b05c1 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_9_7a63bfbb | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_9_40680782 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_9_91ddf9e4 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_9_b33b589a | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_9_de96a73c | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_9_ff4e05ca | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_9_0eda5a2b | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_9_e008d44a | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Handle duplicate elements correctly. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_10_28f530de | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_10_3ef92cac | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_10_b7b5a231 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_10_d417a6d4 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_10_49c33467 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_10_6f4ce0ea | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_10_fa4e1e4c | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_10_e48e99ac | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_10_8dc7c712 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_10_ef7bd5e5 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Return empty result for invalid input. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_11_240d29f3 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_11_86cf7f2f | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_11_06e69e3a | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_11_65c16212 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_11_5a4faf18 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_11_30b94b1b | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_11_1d5f84cf | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_11_1ed43d63 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_11_3e698089 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_11_4f61bfe6 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Minimize memory allocations. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_python_12_daf4bdaa | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Python | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | """ Product Except Self - Brute Force
""" Time and space complexity depends on implementation
def product_except_self(nums):
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Python | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | """ Product Except Self - Prefix Suffix
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | """ Product Except Self - Logarithm
""" Time and space complexity depends on implementation
def product_except_self(nums):
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_javascript_12_94ddd032 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums) {
const n = nums.length;
let result = 0;
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
result = Math.max(result, nums[j]);
}
}
retu... | Approach: Prefix Suffix for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement prefix_suffix solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Logarithm for Product Except Self
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums) {
// TODO: Implement logarithm solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_java_12_e530c712 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Java | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
int n = nums.length;
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = Math... | Approach: Prefix Suffix for Product Except Self
Language: Java | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | Approach: Logarithm for Product Except Self
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public int product_except_self(int[] nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_cpp_12_9456a417 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra sp... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
resul... | Approach: Prefix Suffix for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Ste... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement prefix_suffix solution
int result = 0;
for (int num : nums) {
result += num;
}
re... | Approach: Logarithm for Product Except Self
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexit... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
class Solution {
public:
int product_except_self(vector<int>& nums) {
// TODO: Implement logarithm solution
int result = 0;
for (int num : nums) {
result += num;
}
return res... | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_go_12_3ca2ca10 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Go | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra spa... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
} | Approach: Prefix Suffix for Product Except Self
Language: Go | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
Step... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(nums []int) int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_rust_12_f097b44f | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return resu... | Approach: Prefix Suffix for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
pub fn product_except_self(nums: Vec<i32>) -> i32 {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_typescript_12_fbbc1781 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal e... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return r... | Approach: Prefix Suffix for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correct... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze co... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
function product_except_self(nums: number[]): number {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_csharp_12_fe041f46 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[... | Approach: Prefix Suffix for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze comple... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
public class Solution {
public int ProductExceptSelf(int[] nums) {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_ruby_12_0eeda6a5 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra s... | # Product Except Self - Brute Force
# Time and space complexity depends on implementation
def product_except_self(nums)
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result | Approach: Prefix Suffix for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
St... | # Product Except Self - Prefix Suffix
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result | Approach: Logarithm for Product Except Self
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexi... | # Product Except Self - Logarithm
# Time and space complexity depends on implementation
def product_except_self(nums)
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
product_except_self_swift_12_6da0bf1b | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Support generic types where applicable. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
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