Count the Number of Good Subarrays - Problem

Given an integer array nums and an integer k, return the number of good subarrays of nums.

A subarray arr is good if there are at least k pairs of indices (i, j) such that i < j and arr[i] == arr[j].

A subarray is a contiguous non-empty sequence of elements within an array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,1,4,3,2,2,4], k = 2

› Output: 4

💡 Note: Good subarrays include: [3,1,4,3,2,2] has 2 pairs: (3,3) at indices (0,3) and (2,2) at indices (4,5), so 2≥2✓. [3,1,4,3,2,2,4] has 3 pairs: (3,3), (2,2), and (4,4), so 3≥2✓. Many other subarrays also qualify, totaling 4 good subarrays.

Example 2 — Small Array

$ Input: nums = [1,1,1,1,1], k = 10

› Output: 1

💡 Note: Only the full array [1,1,1,1,1] has enough pairs: C(5,2)=10 pairs of equal elements, which equals k=10.

Example 3 — No Good Subarrays

$ Input: nums = [1,2,3,4], k = 1

› Output: 0

💡 Note: All elements are distinct, so no pairs of equal elements exist. No subarray can have ≥1 pairs.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
1 ≤ k ≤ 10⁹

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 10

The key insight is to use sliding window with frequency counting to efficiently track pairs of equal elements. The optimal approach uses a sliding window that expands to find good subarrays and counts multiple valid subarrays at once. Time: O(n²), Space: O(n)

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n³) Space: O(1)

For each possible subarray, count all pairs of equal elements and check if count >= k. This involves nested loops to generate subarrays and another nested loop to count pairs.

Optimized Sliding Window

⏱️ Time: O(n²) Space: O(n)

Use sliding window technique but count all valid subarrays at once. When we find a good subarray, we can determine how many subarrays starting from current left position are also good, avoiding redundant calculations.

Sliding Window with Frequency Map

⏱️ Time: O(n²) Space: O(n)

Maintain a sliding window where we track frequency of each element. As we expand the window, we calculate how many new pairs are formed. When pairs >= k, count all valid subarrays ending at current position.

Brute Force - Check All Subarrays — Algorithm Steps

Generate all possible subarrays using nested loops
For each subarray, count pairs where arr[i] == arr[j] and i < j
If pair count >= k, increment result counter

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subarrays

Count Pairs

For each subarray, count equal element pairs

Check Condition

If pairs >= k, increment result

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            // Count pairs in subarray nums[i:j+1]
            int pairs = 0;
            for (int x = i; x <= j; x++) {
                for (int y = x + 1; y <= j; y++) {
                    if (nums[x] == nums[y]) {
                        pairs++;
                    }
                }
            }
            
            if (pairs >= k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        if (num != 0 || token[0] == '0') {
            nums[numsSize++] = num;
        }
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n²) to generate subarrays × O(n) to count pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Count the Number of Good Subarrays - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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