Maximum Sum Score of Array

Maximum Sum Score of Array - Problem

You are given a 0-indexed integer array nums of length n.

The sum score of nums at an index i where 0 <= i < n is the maximum of:

The sum of the first i + 1 elements of nums
The sum of the last n - i elements of nums

Return the maximum sum score of nums at any index.

Input & Output

Example 1 — Mixed Positive Numbers

$ Input: nums = [1,4,3,7,4]

› Output: 19

💡 Note: At index 0: left=[1] sum=1, right=[1,4,3,7,4] sum=19, score=19. At index 4: left=[1,4,3,7,4] sum=19, right=[4] sum=4, score=19. Maximum score across all indices is 19.

Example 2 — Single Element

$ Input: nums = [5]

› Output: 5

💡 Note: Only one element, so both left sum and right sum equal 5. Score = max(5,5) = 5.

Example 3 — Negative Numbers

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

› Output: 10

💡 Note: At index 2: left sum = -3+(-5)+2 = -6, right sum = 2+8 = 10, score = max(-6,10) = 10. At index 3: left sum = -3+(-5)+2+8 = 2, right sum = 8, score = max(2,8) = 8. Maximum score is 10.

Constraints

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

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15 🍎 Apple 12

The key insight is to recognize that at each index we need to compare the sum of a prefix with the sum of a suffix. Using prefix sums allows us to calculate both in O(1) time. The optimal approach pre-computes the total sum, then uses a running prefix sum to calculate scores in a single pass. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each index i, calculate the sum of first i+1 elements and sum of last n-i elements by iterating through the array each time. Take the maximum of these two sums and track the overall maximum.

Prefix Sum Optimization

⏱️ Time: O(n) Space: O(n)

Build prefix sum array where prefix[i] contains sum of elements from 0 to i. For each position, left sum is prefix[i] and right sum is total_sum - prefix[i] + nums[i]. This eliminates the need to recalculate sums.

Space-Optimized Single Pass

⏱️ Time: O(n) Space: O(1)

Instead of storing all prefix sums, maintain running prefix sum and calculate suffix sum using total_sum - prefix_sum + current_element. This achieves O(n) time with O(1) extra space by computing total sum first, then processing elements in single pass.

Brute Force — Algorithm Steps

For each index i from 0 to n-1
Calculate sum of elements from 0 to i
Calculate sum of elements from i to n-1
Take maximum of the two sums
Track overall maximum score

Visualization

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Step-by-Step Walkthrough

Position 0

Calculate sum[0:1] and sum[0:5] separately

Position 1

Recalculate sum[0:2] and sum[1:5] from scratch

Continue

Repeat for all positions, finding maximum score

Code -

solution.c — C

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

long long solution(int* nums, int n) {
    long long maxScore = LLONG_MIN;
    
    for (int i = 0; i < n; i++) {
        // Sum of first i+1 elements
        long long leftSum = 0;
        for (int j = 0; j <= i; j++) {
            leftSum += nums[j];
        }
        
        // Sum of last n-i elements
        long long rightSum = 0;
        for (int j = i; j < n; j++) {
            rightSum += nums[j];
        }
        
        long long score = (leftSum > rightSum) ? leftSum : rightSum;
        maxScore = (score > maxScore) ? score : maxScore;
    }
    
    return maxScore;
}

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[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array
    int nums[10000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    long long result = solution(nums, n);
    printf("%lld\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n positions, we sum up to n elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for sums and maximum

✓ Linear Space

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Input & Output

Constraints

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Common Approaches

Brute Force — Algorithm Steps

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Code -

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