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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
merge_intervals_python_15_b28e1085 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_15_0a459f5a | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_15_328683d5 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_15_1c457110 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_15_22ad256e | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_15_60df7484 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_15_69cfa9d6 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_15_22839a63 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_15_0513535b | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_15_c83225ad | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Handle overflow for large values. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_16_2eaac8e1 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_16_3d5d71d6 | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_16_5318ab22 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_16_f4b50495 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_16_7ee381b7 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_16_3e3a2e86 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_16_90a6820f | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_16_0419362b | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_16_87db9e05 | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_16_908bf80a | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Use tail recursion where possible. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_17_890151d4 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_17_7c3c4d8f | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_17_15d8ae4b | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_17_8287f7b2 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_17_5ea85424 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_17_2bf30c18 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_17_87c5d2f7 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_17_0001c905 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_17_ae49266f | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_17_47b5c625 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Leverage immutable data structures. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_18_397c682c | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_18_6aa53919 | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_18_72949b12 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_18_cf888ae4 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_18_812b8531 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_18_491f9c62 | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_18_9ae1cf47 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_18_b0a64176 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_18_d8eb39ea | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_18_938c8558 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Apply early termination for sorted input. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_19_b0e50424 | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_19_33eb7a72 | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_19_dcbe11a2 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_19_64672ddf | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_19_58c5ef67 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_19_bdf1bf3d | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_19_d177bbbf | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_19_e6dbc7b8 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_19_f1932e60 | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_19_640b6b60 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Use bit manipulation for space optimization. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_20_aa2d66ed | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_20_4b0193ee | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_20_ed1f6a3e | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_20_ccef73f5 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_20_983a013b | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_20_39ad3c7d | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_20_ffe92701 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_20_d2633958 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_20_dd2b1ec4 | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_20_d44a32a8 | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Implement using functional programming style. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_python_21_9f1c95ca | algorithms | arrays | medium | python | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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 spa... | """ Merge Intervals - Brute Force
""" Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | """ Merge Intervals - Sorting Greedy
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Python | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | """ Merge Intervals - Heap
""" Time and space complexity depends on implementation
def merge_intervals(nums):
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_javascript_21_01f7df8d | algorithms | arrays | medium | javascript | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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 resul... | Approach: Sorting Greedy for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement sorting_greedy solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | Approach: Heap for Merge Intervals
Language: Javascript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums) {
// TODO: Implement heap solution
let result = 0;
for (const num of nums) {
result += num;
}
return result;
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_java_21_1089b910 | algorithms | arrays | medium | java | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(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(res... | Approach: Sorting Greedy for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
... | Approach: Heap for Merge Intervals
Language: Java | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public int merge_intervals(int[] nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_cpp_21_b8bbe467 | algorithms | arrays | medium | cpp | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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.... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(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(... | Approach: Sorting Greedy for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gre... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement sorting_greedy solution
int result = 0;
for (int num : nums) {
result += num;
}
return r... | Approach: Heap for Merge Intervals
Language: Cpp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
class Solution {
public:
int merge_intervals(vector<int>& nums) {
// TODO: Implement heap solution
int result = 0;
for (int num : nums) {
result += num;
}
return result;
}
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_go_21_898952d3 | algorithms | arrays | medium | go | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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.
... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gree... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Go | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-heap ... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(nums []int) int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_rust_21_60f5a8bc | algorithms | arrays | medium | rust | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
pub fn merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Rust | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
pub fn merge_intervals(nums: Vec<i32>) -> i32 {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_typescript_21_256d2957 | algorithms | arrays | medium | typescript | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
function merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-p... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Typescript | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, m... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
function merge_intervals(nums: number[]): number {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_csharp_21_edbbcea0 | algorithms | arrays | medium | csharp | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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 spa... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(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])
... | Approach: Sorting Greedy for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass ... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Csharp | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-h... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
public class Solution {
public int MergeIntervals(int[] nums) {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_ruby_21_7cafd52a | algorithms | arrays | medium | ruby | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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... | # Merge Intervals - Brute Force
# Time and space complexity depends on implementation
def merge_intervals(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: Sorting Greedy for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass gr... | # Merge Intervals - Sorting Greedy
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result | Approach: Heap for Merge Intervals
Language: Ruby | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-hea... | # Merge Intervals - Heap
# Time and space complexity depends on implementation
def merge_intervals(nums)
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
merge_intervals_swift_21_ca6327aa | algorithms | arrays | medium | swift | Given a collection of intervals, merge all overlapping intervals. Add memoization to avoid recomputation. | Approach: Brute Force for Merge Intervals
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 spac... | // Merge Intervals - Brute Force
// Time and space complexity depends on implementation
func merge_intervals(_ 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: Sorting Greedy for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Sort items by a key criteria to enable greedy processing.
Step 2: Process items in sorted order, maintaining running state.
Step 3: Merge or combine items based on overlap/relationship.
Step 4: The sorting enables a single-pass g... | // Merge Intervals - Sorting Greedy
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement sorting_greedy solution
result = 0
for num in nums:
result += num
return result
} | Approach: Heap for Merge Intervals
Language: Swift | Difficulty: medium
Step 1: Use a heap (priority queue) for efficient min/max access.
Step 2: Heaps provide O(log N) insert/extract, O(1) peek.
Step 3: Min-heap for smallest, max-heap for largest element access.
Step 4: For median: maintain max-heap for lower, min-he... | // Merge Intervals - Heap
// Time and space complexity depends on implementation
func merge_intervals(_ nums: [Int]) -> Int {
# TODO: Implement heap solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "sorting_greedy",
"approach_3_type": "heap_based",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N log N)",
"time_complexity_3": "O(N log K)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"merge_intervals... |
product_except_self_python_0_3f9106b2 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. | 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_0_ac9ca0f4 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. | 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_0_d1d424e1 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. | 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_0_0eaa0f68 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. | 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_0_8ae39968 | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. | 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_0_38fc302b | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. | 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_0_ec275c33 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. | 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_0_3bf776cd | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. | 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_0_6be7fb32 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. | 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_0_c55b271f | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. | 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_1_9d823fac | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_d1863757 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_97f23082 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_67dda142 | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_d442626f | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_28dfee91 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_331e989d | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_016c41c8 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_712b1033 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_1_f8fbba4c | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Optimize for large input sizes up to 10^6. | 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_2_4f098f97 | algorithms | arrays | medium | python | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_72de6b51 | algorithms | arrays | medium | javascript | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_b21a8811 | algorithms | arrays | medium | java | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_8b98461b | algorithms | arrays | medium | cpp | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_a5f63dda | algorithms | arrays | medium | go | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_f4571b55 | algorithms | arrays | medium | rust | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_25875a31 | algorithms | arrays | medium | typescript | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_cbb4b7d7 | algorithms | arrays | medium | csharp | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_ee783325 | algorithms | arrays | medium | ruby | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | 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_2_ccdaf6ef | algorithms | arrays | medium | swift | Return an array where each element is the product of all other elements. Use iterative approach to avoid recursion limits. | Approach: Brute Force for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Start with the most straightforward approach: iterate through all possible candidates.
Step 2: Check every valid combination against the problem requirements.
Step 3: This gives higher time complexity but requires minimal extra ... | // Product Except Self - Brute Force
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
n = len(nums)
result = 0
for i in range(n):
for j in range(i, n):
# Process subproblem
result = max(result, nums[j])
return result
... | Approach: Prefix Suffix for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Compute prefix products and suffix products separately.
Step 2: For each position, result is prefix[i-1] * suffix[i+1].
Step 3: Two passes: one for prefix, one for suffix.
Step 4: Avoids division and handles zeros correctly.
S... | // Product Except Self - Prefix Suffix
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement prefix_suffix solution
result = 0
for num in nums:
result += num
return result
} | Approach: Logarithm for Product Except Self
Language: Swift | Difficulty: medium
Step 1: Analyze the problem requirements and constraints carefully.
Step 2: Choose logarithm as the strategy.
Step 3: Implement the solution step by step, handling edge cases.
Step 4: Verify correctness with test cases and analyze complex... | // Product Except Self - Logarithm
// Time and space complexity depends on implementation
func product_except_self(_ nums: [Int]) -> Int {
# TODO: Implement logarithm solution
result = 0
for num in nums:
result += num
return result
} | {
"approach_1_type": "brute_force",
"approach_2_type": "prefix_suffix_arrays",
"approach_3_type": "logarithm",
"time_complexity_1": "O(N^2)",
"time_complexity_2": "O(N)",
"time_complexity_3": "O(N)",
"tags": [
"array",
"iteration",
"search",
"difficulty-medium",
"product_except_self"
... |
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