Count Subarrays Where Max Element Appears at Least K Times - Problem

You are given an integer array nums and a positive integer k.

Return the number of subarrays where the maximum element of nums appears at least k times in that subarray.

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

Input & Output

Example 1 — Basic Case

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

› Output: 6

💡 Note: Max element is 3. Subarrays with at least 2 occurrences of 3: [3,2,3], [3,2,3,3], [2,3,3], [3,3], [3,2,3,3] (starting from index 1), and [3,3] (starting from index 3). Total: 6 subarrays.

Example 2 — Minimum Size

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

› Output: 0

💡 Note: Max element is 4, appears only once. Since k=3 but max element appears only 1 time, no subarray can have the max element appearing at least 3 times.

Example 3 — All Same Elements

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

› Output: 3

💡 Note: Max element is 3. Valid subarrays: [3,3] (indices 0-1), [3,3,3] (indices 0-2), and [3,3] (indices 1-2). Total: 3 subarrays.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁶
1 ≤ k ≤ nums.length

Visualization

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Asked in

G Google 25 A Amazon 18 M Microsoft 15

The key insight is to use a sliding window approach to efficiently count subarrays where the maximum element appears at least k times. Best approach is sliding window with Time: O(n), Space: O(1).

Common Approaches

✓ Math

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate all possible subarrays, find the maximum element in each, then count how many times it appears. If count >= k, increment result.

Sliding Window

⏱️ Time: O(n) Space: O(1)

Maintain a sliding window where we track the count of maximum elements. For each right position, find the leftmost position where max count >= k, then all subarrays ending at right position starting from that position or later are valid.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

long long solution(int* nums, int numsSize, int k) {
    if (numsSize == 0) return 0;
    
    // Find the global maximum
    int maxVal = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] > maxVal) {
            maxVal = nums[i];
        }
    }
    
    long long count = 0;
    
    // Check all possible subarrays
    for (int i = 0; i < numsSize; i++) {
        int maxCount = 0;
        for (int j = i; j < numsSize; j++) {
            // If current element is the global max, increment count
            if (nums[j] == maxVal) {
                maxCount++;
            }
            
            // If we have at least k occurrences of max element
            if (maxCount >= k) {
                count++;
            }
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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