Maximum Subarray Sum After One Operation

Maximum Subarray Sum After One Operation - Problem

You are given an integer array nums. You must perform exactly one operation where you can replace one element nums[i] with nums[i] * nums[i].

Return the maximum possible subarray sum after exactly one operation. The subarray must be non-empty.

Note: A subarray is a contiguous part of an array.

Input & Output

Example 1 — Basic Case

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

› Output: 17

💡 Note: Square -4 to get 16, then subarray [-1,16] gives sum 15, but subarray [16] alone gives 16, and we can extend to get 17 by including previous elements optimally

Example 2 — Single Element

$ Input: nums = [1]

› Output: 1

💡 Note: Only one element, must square it: 1² = 1

Example 3 — All Positive

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

› Output: 22

💡 Note: Square the largest element 4 to get 16, total subarray sum: 1+2+3+16 = 22

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 45 f Facebook 32 M Microsoft 28 a Amazon 25

The key insight is to use dynamic programming to track two states: maximum subarray sum without using the square operation and maximum subarray sum with exactly one square operation used. The optimal approach uses Enhanced Kadane's Algorithm with DP state transitions. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each position, square that element and calculate the maximum subarray sum using Kadane's algorithm. Take the maximum across all possibilities.

Enhanced Kadane's Algorithm

⏱️ Time: O(n) Space: O(1)

Use dynamic programming to track two states: maximum subarray ending here without squaring any element, and maximum subarray ending here with exactly one element squared.

Brute Force — Algorithm Steps

Step 1: For each index i, create a copy of array with nums[i] squared
Step 2: Apply Kadane's algorithm to find maximum subarray sum
Step 3: Return the maximum sum across all attempts

Visualization

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Step-by-Step Walkthrough

Try Position 0

Square first element and run Kadane's

Try Position 1

Square second element and run Kadane's

Continue

Test all positions and take maximum

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int maxSum = INT_MIN;
    
    // Try squaring each element
    for (int i = 0; i < n; i++) {
        // Create modified array
        int* modified = malloc(n * sizeof(int));
        memcpy(modified, nums, n * sizeof(int));
        modified[i] = nums[i] * nums[i];
        
        // Apply Kadane's algorithm
        int currentSum = modified[0];
        int bestSum = modified[0];
        
        for (int j = 1; j < n; j++) {
            currentSum = (modified[j] > currentSum + modified[j]) ? modified[j] : currentSum + modified[j];
            bestSum = (currentSum > bestSum) ? currentSum : bestSum;
        }
        
        maxSum = (bestSum > maxSum) ? bestSum : maxSum;
        free(modified);
    }
    
    return maxSum;
}

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[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Outer loop tries n positions, inner Kadane's algorithm takes O(n) for each

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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