Find the Power of K-Size Subarrays I - Practice Coding Problems

Find the Power of K-Size Subarrays I - 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 Case

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

› Output: [3,4]

💡 Note: First subarray [1,2,3] is consecutive ascending, power = 3. Second subarray [2,3,4] is consecutive ascending, power = 4.

Example 2 — Non-consecutive Elements

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

› Output: [-1,-1]

💡 Note: All elements are the same (not ascending), so no subarray has power. Both [2,2,2,2] subarrays return -1.

Example 3 — Gap in Sequence

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

› Output: [-1,-1,-1,-1]

💡 Note: No 2-element subarray has consecutive ascending elements: [3,2], [2,3], [3,2], [2,3] all have gaps or are not ascending.

Constraints

1 ≤ n ≤ 500
1 ≤ nums[i] ≤ 10⁵
1 ≤ k ≤ n

Visualization

Tap to expand

Understanding the Visualization

Input Array

Given array nums and window size k

Check Windows

For each k-size subarray, verify if consecutive ascending

Output Powers

Return max element if valid, -1 otherwise

Key Takeaway

🎯 Key Insight: Use sliding window to efficiently track consecutive count and avoid redundant checks

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that we need to check if k consecutive elements form an ascending sequence. The brute force approach checks each subarray individually in O(n×k) time. The sliding window approach optimizes this to O(n) by tracking consecutive count as we slide. Best approach is sliding window with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n×k)	O(1)	Check each k-size subarray individually for consecutive ascending order
Sliding Window with Consecutive Tracking	O(n)	O(1)	Track consecutive count as we slide the window, avoiding redundant checks

Brute Force — Algorithm Steps

Step 1: For each starting position i from 0 to n-k
Step 2: Check if subarray[i..i+k-1] has consecutive ascending elements
Step 3: If valid, add max element to result, otherwise add -1

Visualization

Tap to expand

Step-by-Step Walkthrough

Check First Subarray

Verify if elements are consecutive ascending

Check Next Subarray

Move window and repeat check

Record Power

Add max element or -1 to result

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* result = (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; j < i + k - 1; j++) {
            if (nums[j + 1] != nums[j] + 1) {
                isValid = 0;
                break;
            }
        }
        
        if (isValid) {
            result[i] = nums[i + k - 1];  // Maximum element
        } else {
            result[i] = -1;
        }
    }
    
    return result;
}

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) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    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 for consecutiveness

✓ Linear Growth

Space Complexity

O(1)

Only using variables for iteration, result array doesn't count as extra space

✓ Linear Space

12.5K Views

Medium Frequency

~25 min Avg. Time

456 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 — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler