Find the Middle Index in Array

Find the Middle Index in Array - Problem

Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones).

A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].

If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0.

Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.

Input & Output

Example 1 — Basic Case

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

› Output: 0

💡 Note: At index 0: left sum = 0 (no elements), right sum = 1 + (-1) = 0. Since 0 = 0, return 0.

Example 2 — Middle Balance

$ Input: nums = [2,3,1,0,4]

› Output: -1

💡 Note: No index satisfies the condition. At index 0: left=0, right=8. At index 1: left=2, right=5. At index 2: left=5, right=4. At index 3: left=6, right=4. At index 4: left=6, right=0.

Example 3 — No Balance Point

$ Input: nums = [1,7,3,6,5,6]

› Output: 3

💡 Note: At index 3: left sum = 1 + 7 + 3 = 11, right sum = 5 + 6 = 11. Since 11 = 11, return 3.

Constraints

1 ≤ nums.length ≤ 10⁴
-1000 ≤ nums[i] ≤ 1000

Visualization

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Asked in

a Amazon 45 M Microsoft 32 f Facebook 28 G Google 21

The key insight is to use the total sum to calculate right sum efficiently. Instead of recalculating sums repeatedly, compute total sum once, then use rightSum = totalSum - leftSum - currentElement. Best approach is prefix sum with Time: O(n), Space: O(1).

Common Approaches

✓ Memoization

⏱️ Time: N/A Space: N/A

Dp 2d

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n²) Space: O(1)

For every possible middle index, calculate the sum of all elements to its left and all elements to its right. Compare these sums and return the first index where they are equal.

Prefix Sum Optimization

⏱️ Time: O(n) Space: O(1)

First calculate the total sum of all elements. Then iterate through the array maintaining a running left sum. The right sum can be calculated as total - left_sum - current_element.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int findMiddleIndex(int* nums, int numsSize) {
    if (numsSize == 0) return -1;
    
    // Calculate total sum
    int totalSum = 0;
    for (int i = 0; i < numsSize; i++) {
        totalSum += nums[i];
    }
    
    int leftSum = 0;
    
    // Check each position as potential middle index
    for (int i = 0; i < numsSize; i++) {
        // Right sum = total - left - current element
        int rightSum = totalSum - leftSum - nums[i];
        
        if (leftSum == rightSum) {
            return i;
        }
        
        // Update left sum for next iteration
        leftSum += nums[i];
    }
    
    return -1;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    
    // Skip to '['
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    // Parse numbers
    while (*p && *p != ']') {
        // Skip whitespace and commas
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        // Parse number (handle negative numbers)
        char* endp;
        arr[(*size)++] = (int)strtol(p, &endp, 10);
        p = endp;
    }
}

int main() {
    static int nums[1000];  // Use static to avoid stack overflow
    int numsSize = 0;
    
    char line[10000];
    if (fgets(line, sizeof(line), stdin) == NULL) {
        return -1;
    }
    
    parseArray(line, nums, &numsSize);
    
    int result = findMiddleIndex(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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