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_13_bd853002 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_86a7a09a | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_eb555d45 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_1cad6f14 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_9aa9de7e | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_82270e9a | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_a6c2001f | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_04b65514 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_1d7dead6 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_13_c30eb081 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Document time and space complexity. | 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_14_dfce2b64 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_55102e97 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_a2d94677 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_bb8f615d | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_d2cd768d | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_b156719f | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_fb147614 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_a120a329 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_43377921 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_14_9a0115f6 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Include unit tests for verification. | 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_15_1134c1a7 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_77255bcf | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_a9f2fce2 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_01e8078d | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_9d0fbad5 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_ef4f7821 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_b349c144 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_2cb6280d | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_fc9091f1 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_15_7a120635 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Handle overflow for large values. | 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_16_4a4bd6b9 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_4c925513 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_4f3bdcbd | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_bf7da9c9 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_24b229d1 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_5396b4b8 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_4eed9a2b | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_42eb0e96 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_1a8afd25 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_16_28513af3 | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Use tail recursion where possible. | 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_17_7ac4f4b7 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_12eaaaad | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_b59a02c2 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_45058a6a | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_2ee24e4e | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_dbfbf43f | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_9d1e4c11 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_8f578c2c | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_a90a466e | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_17_323be38d | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Leverage immutable data structures. | 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_18_3b2b9e63 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_c97f2c59 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_13f95a26 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_5173b037 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_335e4ebd | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_8f4ca639 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_2ef03d39 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_0d351f82 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_9b234a1f | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_18_ab61c26e | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Apply early termination for sorted 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_19_e249f887 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_0edd6ca2 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_7441a81c | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_a40afcef | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_8d1fdadb | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_06f1b7e9 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_66347930 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_72eb6834 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_a5e32e7c | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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_19_8bd5d10e | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Use bit manipulation for space optimization. | 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"
... |
rotate_array_python_0_0b122f25 | algorithms | arrays | easy | python | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Python | Difficulty: easy
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 space.
S... | """ Rotate Array - Brute Force
""" Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Python | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | """ Rotate Array - Reverse
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Python | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | """ Rotate Array - Cyclic Replacement
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_javascript_0_8fd615f9 | algorithms | arrays | easy | javascript | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Javascript | Difficulty: easy
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 spac... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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]);
}
}
return result;
} | Approach: Reverse for Rotate Array
Language: Javascript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement reverse solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Cyclic Replacement for Rotate Array
Language: Javascript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and a... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement cyclic_replacement solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_java_0_c6bcb503 | algorithms | arrays | easy | java | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Java | Difficulty: easy
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 space.
Ste... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(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.max(result, n... | Approach: Reverse for Rotate Array
Language: Java | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Java | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_cpp_0_4d23d24d | algorithms | arrays | easy | cpp | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Cpp | Difficulty: easy
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 space.
Step... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(result... | Approach: Reverse for Rotate Array
Language: Cpp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Cpp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze ... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_go_0_ab8d5185 | algorithms | arrays | easy | go | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Go | Difficulty: easy
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 space.
Step ... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(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: Reverse for Rotate Array
Language: Go | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Go | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze c... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_rust_0_f669cb50 | algorithms | arrays | easy | rust | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Rust | Difficulty: easy
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 space.
Ste... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
pub fn rotate_array(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 result
} | Approach: Reverse for Rotate Array
Language: Rust | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Rust | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_typescript_0_28e2d20c | algorithms | arrays | easy | typescript | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Typescript | Difficulty: easy
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 spac... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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 result | Approach: Reverse for Rotate Array
Language: Typescript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Typescript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and a... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_csharp_0_ff04a8c5 | algorithms | arrays | easy | csharp | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Csharp | Difficulty: easy
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 space.
S... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
retur... | Approach: Reverse for Rotate Array
Language: Csharp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Csharp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_ruby_0_109896fe | algorithms | arrays | easy | ruby | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Ruby | Difficulty: easy
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 space.
Ste... | # Rotate Array - Brute Force
# Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Ruby | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | # Rotate Array - Reverse
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Ruby | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | # Rotate Array - Cyclic Replacement
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_swift_0_4e2df9f2 | algorithms | arrays | easy | swift | Rotate the array to the right by k steps. | Approach: Brute Force for Rotate Array
Language: Swift | Difficulty: easy
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 space.
St... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(_ 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: Reverse for Rotate Array
Language: Swift | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Swift | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyz... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_python_1_7bfb1e8b | algorithms | arrays | medium | python | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.... | """ Rotate Array - Brute Force
""" Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | """ Rotate Array - Reverse
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Python | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and ana... | """ Rotate Array - Cyclic Replacement
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_javascript_1_e6f1a192 | algorithms | arrays | medium | javascript | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 extra sp... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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]);
}
}
return result;
} | Approach: Reverse for Rotate Array
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement reverse solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Cyclic Replacement for Rotate Array
Language: Javascript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement cyclic_replacement solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_java_1_1700c939 | algorithms | arrays | medium | java | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
S... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(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.max(result, n... | Approach: Reverse for Rotate Array
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Java | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_cpp_1_eb91f40b | algorithms | arrays | medium | cpp | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
St... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(result... | Approach: Reverse for Rotate Array
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Cpp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyz... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_go_1_050ff0f7 | algorithms | arrays | medium | go | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
Ste... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(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: Reverse for Rotate Array
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Go | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_rust_1_bf801e91 | algorithms | arrays | medium | rust | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
S... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
pub fn rotate_array(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 result
} | Approach: Reverse for Rotate Array
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Rust | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_typescript_1_56430852 | algorithms | arrays | medium | typescript | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 extra sp... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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 result | Approach: Reverse for Rotate Array
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Typescript | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_csharp_1_4476d084 | algorithms | arrays | medium | csharp | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
retur... | Approach: Reverse for Rotate Array
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Csharp | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and ana... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_ruby_1_9a74d426 | algorithms | arrays | medium | ruby | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
S... | # Rotate Array - Brute Force
# Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | # Rotate Array - Reverse
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Ruby | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | # Rotate Array - Cyclic Replacement
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_swift_1_df34b39b | algorithms | arrays | medium | swift | Rotate the array to the right by k steps. Optimize for large input sizes up to 10^6. | Approach: Brute Force for Rotate Array
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 space.
... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(_ 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: Reverse for Rotate Array
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and anal... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"rotate_array"
],
"esti... |
rotate_array_python_2_2fd78c0a | algorithms | arrays | easy | python | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Python | Difficulty: easy
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 space.
S... | """ Rotate Array - Brute Force
""" Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Python | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | """ Rotate Array - Reverse
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Python | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | """ Rotate Array - Cyclic Replacement
""" Time and space complexity depends on implementation
def rotate_array(nums):
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_javascript_2_ae1c1fb3 | algorithms | arrays | easy | javascript | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Javascript | Difficulty: easy
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 spac... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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]);
}
}
return result;
} | Approach: Reverse for Rotate Array
Language: Javascript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement reverse solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Cyclic Replacement for Rotate Array
Language: Javascript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and a... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums) {
// TODO: Implement cyclic_replacement solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_java_2_52d212ed | algorithms | arrays | easy | java | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Java | Difficulty: easy
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 space.
Ste... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(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.max(result, n... | Approach: Reverse for Rotate Array
Language: Java | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Java | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public int rotate_array(int[] nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_cpp_2_a99ad6c8 | algorithms | arrays | easy | cpp | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Cpp | Difficulty: easy
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 space.
Step... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
int n = nums.size();
int result = 0;
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
result = max(result... | Approach: Reverse for Rotate Array
Language: Cpp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement reverse solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | Approach: Cyclic Replacement for Rotate Array
Language: Cpp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze ... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
class Solution {
public:
int rotate_array(vector<int>& nums) {
// TODO: Implement cyclic_replacement solution
int result = 0;
for (int num : nums) {
result += num;
}
return... | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_go_2_2af327af | algorithms | arrays | easy | go | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Go | Difficulty: easy
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 space.
Step ... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(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: Reverse for Rotate Array
Language: Go | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Go | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze c... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(nums []int) int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_rust_2_f514056d | algorithms | arrays | easy | rust | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Rust | Difficulty: easy
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 space.
Ste... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
pub fn rotate_array(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 result
} | Approach: Reverse for Rotate Array
Language: Rust | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Rust | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
pub fn rotate_array(nums: Vec<i32>) -> i32 {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_typescript_2_606dc436 | algorithms | arrays | easy | typescript | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Typescript | Difficulty: easy
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 spac... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
function rotate_array(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 result | Approach: Reverse for Rotate Array
Language: Typescript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Typescript | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and a... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
function rotate_array(nums: number[]): number {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_csharp_2_4708154e | algorithms | arrays | easy | csharp | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Csharp | Difficulty: easy
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 space.
S... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
retur... | Approach: Reverse for Rotate Array
Language: Csharp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Csharp | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analy... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
public class Solution {
public int RotateArray(int[] nums) {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_ruby_2_ec78fc48 | algorithms | arrays | easy | ruby | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Ruby | Difficulty: easy
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 space.
Ste... | # Rotate Array - Brute Force
# Time and space complexity depends on implementation
def rotate_array(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: Reverse for Rotate Array
Language: Ruby | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | # Rotate Array - Reverse
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result | Approach: Cyclic Replacement for Rotate Array
Language: Ruby | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze... | # Rotate Array - Cyclic Replacement
# Time and space complexity depends on implementation
def rotate_array(nums)
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
rotate_array_swift_2_2e3111ae | algorithms | arrays | easy | swift | Rotate the array to the right by k steps. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Rotate Array
Language: Swift | Difficulty: easy
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 space.
St... | // Rotate Array - Brute Force
// Time and space complexity depends on implementation
func rotate_array(_ 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: Reverse for Rotate Array
Language: Swift | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose reverse as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complexity. | // Rotate Array - Reverse
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement reverse solution
result = 0
for num in nums:
result += num
return result
} | Approach: Cyclic Replacement for Rotate Array
Language: Swift | Difficulty: easy
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose cyclic replacement as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyz... | // Rotate Array - Cyclic Replacement
// Time and space complexity depends on implementation
func rotate_array(_ nums: [Int]) -> Int {
# TODO: Implement cyclic_replacement solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "reverse",
"approach_3_type": "cyclic_replacement",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-easy",
"rotate_array"
],
"estima... |
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