Maximum Average Subarray I

Maximum Average Subarray I - Problem

You are given an integer array nums consisting of n elements, and an integer k.

Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value.

Any answer with a calculation error less than 10^-5 will be accepted.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,12,-5,-6,50,3], k = 4

› Output: 12.75000

💡 Note: Maximum average subarray is [12,-5,-6,50] with sum 51 and average 51/4 = 12.75

Example 2 — Small Array

$ Input: nums = [5], k = 1

› Output: 5.00000

💡 Note: Only one element, so maximum average is 5.0

Example 3 — All Negative

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

› Output: -1.50000

💡 Note: Best subarray is [-1,-2] with average -3/2 = -1.5

Constraints

n == nums.length
1 ≤ k ≤ n ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 32 a Amazon 28 M Microsoft 15

The sliding window technique is the optimal approach. Instead of recalculating the sum for each window, we calculate the first window sum and then slide by removing the leftmost element and adding the rightmost element. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Calculate Every Window	O(n×k)	O(1)	Calculate sum and average for every possible k-length window
Sliding Window - Optimal Solution	O(n)	O(1)	Use sliding window technique to avoid recalculating overlapping elements

Brute Force - Calculate Every Window — Algorithm Steps

Iterate through all possible starting positions (0 to n-k)
For each position, calculate sum of k elements
Track maximum average found so far

Visualization

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Step-by-Step Walkthrough

Window 1

Calculate sum of first k elements

Window 2

Calculate sum of next k elements (overlapping with window 1)

Continue

Repeat for all possible windows and find maximum average

Code -

solution.c — C

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

double solution(int* nums, int numsSize, int k) {
    double maxAverage = -DBL_MAX;
    
    for (int i = 0; i <= numsSize - k; i++) {
        int currentSum = 0;
        for (int j = i; j < i + k; j++) {
            currentSum += nums[j];
        }
        double currentAverage = (double) currentSum / k;
        if (currentAverage > maxAverage) {
            maxAverage = currentAverage;
        }
    }
    
    return maxAverage;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n×k)

Outer loop runs (n-k+1) times, inner loop runs k times for each window

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track sum and maximum average

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Calculate Every Window — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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