Array Sum and Average - Problem

Given an array of numbers, calculate and return an array containing the sum, average, minimum, and maximum values from the input array.

The output array should contain exactly 4 elements in this order:

[0]: Sum of all elements
[1]: Average of all elements (as a decimal)
[2]: Minimum element
[3]: Maximum element

Note: The average should be calculated as a floating-point number, not rounded to an integer.

Input & Output

Example 1 — Basic Array

$ Input: nums = [1,5,3,9,2]

› Output: [20,4.0,1,9]

💡 Note: Sum: 1+5+3+9+2 = 20, Average: 20/5 = 4.0, Min: 1, Max: 9

Example 2 — Negative Numbers

$ Input: nums = [-2,4,-1,7]

› Output: [8,2.0,-2,7]

💡 Note: Sum: -2+4+(-1)+7 = 8, Average: 8/4 = 2.0, Min: -2, Max: 7

Example 3 — Single Element

$ Input: nums = [42]

› Output: [42,42.0,42,42]

💡 Note: With one element, sum, average, min, and max are all the same value: 42

Constraints

1 ≤ nums.length ≤ 10⁴
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to calculate all statistics (sum, minimum, maximum) in a single pass through the array rather than making separate loops. The optimal approach tracks running sum, min, and max simultaneously, then computes the average. Time: O(n), Space: O(1)

Common Approaches

✓ Multiple Pass Approach

⏱️ Time: O(n) Space: O(1)

Use four separate passes through the array to calculate sum, average, minimum, and maximum values independently. This is straightforward but inefficient as it processes the same data multiple times.

Single Pass Approach

⏱️ Time: O(n) Space: O(1)

Traverse the array only once, calculating sum, minimum, and maximum simultaneously. Then compute the average using the sum and array length. This is much more efficient than multiple passes.

Multiple Pass Approach — Algorithm Steps

First pass: Loop through array to calculate sum
Second pass: Loop through array to find minimum value
Third pass: Loop through array to find maximum value
Calculate average by dividing sum by array length

Visualization

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Step-by-Step Walkthrough

Pass 1: Sum

Loop through array to calculate total sum

Pass 2: Min

Loop through array to find minimum value

Pass 3: Max

Loop through array to find maximum value

Calculate Average

Divide sum by array length

Code -

solution.c — C

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

void solution(int* nums, int size, double* result) {
    // Calculate sum
    int total = 0;
    for (int i = 0; i < size; i++) {
        total += nums[i];
    }
    
    // Calculate minimum
    int minimum = nums[0];
    for (int i = 0; i < size; i++) {
        if (nums[i] < minimum) {
            minimum = nums[i];
        }
    }
    
    // Calculate maximum
    int maximum = nums[0];
    for (int i = 0; i < size; i++) {
        if (nums[i] > maximum) {
            maximum = nums[i];
        }
    }
    
    // Calculate average
    double average = (double)total / size;
    
    result[0] = total;
    result[1] = average;
    result[2] = minimum;
    result[3] = maximum;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[100];
    int count = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        if (num != 0 || strcmp(token, "0") == 0) {
            nums[count++] = num;
        }
        token = strtok(NULL, "[,]");
    }
    
    double result[4];
    solution(nums, count, result);
    
    printf("[%g,%g,%g,%g]\n", result[0], result[1], result[2], result[3]);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Three separate loops through the array, but still linear overall

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to store the results

✓ Linear Space

12.0K Views

Medium Frequency

~8 min Avg. Time

450 Likes

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Array Sum and Average - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Multiple Pass Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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