Find the Power of K-Size Subarrays II - Problem

You are given an array of integers nums of length n and a positive integer k.

The power of an array is defined as:

Its maximum element if all of its elements are consecutive and sorted in ascending order.
-1 otherwise.

You need to find the power of all subarrays of nums of size k.

Return an integer array results of size n - k + 1, where results[i] is the power of nums[i..(i + k - 1)].

Input & Output

Example 1 — Basic Consecutive Sequence

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

› Output: [4]

💡 Note: The subarray [1,2,3,4] has all elements consecutive and sorted in ascending order. The maximum element is 4, so the power is 4.

Example 2 — Mixed Consecutive and Non-consecutive

$ Input: nums = [2,2,2,2,5,5,5,8], k = 3

› Output: [-1,-1,-1,-1,-1,-1]

💡 Note: No subarray of size 3 has consecutive ascending elements. All duplicates or gaps prevent consecutive sequences.

Example 3 — Partial Consecutive Sequence

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

› Output: [-1,3,-1,3,-1]

💡 Note: Windows are [3,2]→-1 (not ascending), [2,3]→3 (consecutive ascending), [3,2]→-1, [2,3]→3, [3,2]→-1.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ k ≤ n
1 ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

G Google 25 M Microsoft 18 A Amazon 15

The key insight is to track consecutive elements efficiently while sliding the window. The optimal sliding window approach maintains a count of consecutive elements, avoiding redundant checks. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Check Each Subarray

⏱️ Time: O(n×k) Space: O(1)

For every possible subarray of size k, iterate through its elements to verify they form a consecutive ascending sequence. If they do, return the maximum (last element), otherwise return -1.

Sliding Window with Consecutive Counter

⏱️ Time: O(n) Space: O(1)

Use a sliding window approach where we maintain a counter of consecutive elements. As we slide the window, we efficiently update this counter instead of rechecking all elements.

Brute Force - Check Each Subarray — Algorithm Steps

Iterate through all possible starting positions
For each subarray, check if elements are consecutive and ascending
Return max element if valid, -1 otherwise

Visualization

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Step-by-Step Walkthrough

Check First Subarray

Verify if elements are consecutive ascending

Check Next Subarray

Move window and check again

Continue

Repeat for all positions

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int k, int* returnSize) {
    *returnSize = numsSize - k + 1;
    int* results = (int*)malloc(*returnSize * sizeof(int));
    
    for (int i = 0; i <= numsSize - k; i++) {
        // Check if subarray nums[i:i+k] is consecutive and ascending
        int isValid = 1;
        for (int j = i + 1; j < i + k; j++) {
            if (nums[j] != nums[j-1] + 1) {
                isValid = 0;
                break;
            }
        }
        
        if (isValid) {
            results[i] = nums[i + k - 1]; // Maximum element
        } else {
            results[i] = -1;
        }
    }
    
    return results;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array - handle proper JSON format
    int nums[1000];
    int numsSize = 0;
    
    // Skip '[' and find numbers
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++;
    if (*ptr == '[') ptr++;
    
    char* start = ptr;
    while (*ptr && *ptr != ']') {
        if (*ptr == ',' || *ptr == ']') {
            *ptr = '\0';
            if (start != ptr) {
                nums[numsSize++] = atoi(start);
            }
            start = ptr + 1;
        }
        ptr++;
    }
    // Handle last number before ']'
    if (start < ptr && *start >= '0' && *start <= '9') {
        nums[numsSize++] = atoi(start);
    }
    
    int k;
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(nums, numsSize, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×k)

For each of n-k+1 subarrays, we check k elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Find the Power of K-Size Subarrays II - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Subarray — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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