Number of Good Pairs

Number of Good Pairs - Problem

Given an array of integers nums, return the number of good pairs.

A pair (i, j) is called good if nums[i] == nums[j] and i < j.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,1,1,3]

› Output: 4

💡 Note: Good pairs are: (0,3), (0,4), (3,4), (2,5). nums[0]==nums[3]=1, nums[0]==nums[4]=1, nums[3]==nums[4]=1, nums[2]==nums[5]=3.

Example 2 — No Pairs

$ Input: nums = [1,2,3]

› Output: 0

💡 Note: All elements are unique, so no good pairs exist where nums[i] == nums[j] with i < j.

Example 3 — All Same

$ Input: nums = [1,1,1,1]

› Output: 6

💡 Note: All pairs are good: (0,1), (0,2), (0,3), (1,2), (1,3), (2,3). With 4 identical elements, we get 4×3/2 = 6 pairs.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100

Visualization

Tap to expand

Asked in

LC LeetCode 15 G Google 8

The key insight is that for any number appearing n times, it forms n×(n-1)/2 good pairs. The optimal approach uses a hash map to count pairs as we iterate, achieving O(n) time and O(n) space complexity.

Common Approaches

✓ Brute Force - Nested Loops

⏱️ Time: O(n²) Space: O(1)

Use two nested loops to examine all possible pairs. For each element at index i, check all elements at indices j > i. If nums[i] equals nums[j], increment the count.

Frequency Count - Hash Map

⏱️ Time: O(n) Space: O(n)

Use a hash map to count how many times each number appears. For each number that appears n times, it can form n*(n-1)/2 good pairs. Sum up all pairs for each unique number.

One Pass Count - Optimized

⏱️ Time: O(n) Space: O(n)

Instead of counting frequencies first, we count pairs on-the-fly. When we encounter a number, the current count of that number tells us how many new pairs we can form with previously seen occurrences.

Brute Force - Nested Loops — Algorithm Steps

Iterate through each element with index i
For each i, iterate through elements with index j > i
If nums[i] == nums[j], increment good pairs count
Return total count

Visualization

Tap to expand

Step-by-Step Walkthrough

Setup

Initialize count = 0, iterate through all pairs

Compare

For each pair, check if values are equal

Count

Increment count when values match

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] == nums[j]) {
                count++;
            }
        }
    }
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops iterate through all n*(n-1)/2 pairs

✓ Linear Growth

Space Complexity

O(1)

Only uses a constant amount of extra variables

⚡ Linearithmic Space

128.0K Views

Medium Frequency

~10 min Avg. Time

4.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Nested Loops — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler