Minimum Difference in Sums After Removal of Elements

Minimum Difference in Sums After Removal of Elements - Problem

You are given a 0-indexed integer array nums consisting of 3 * n elements.

You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:

The first n elements belonging to the first part and their sum is sumfirst.
The next n elements belonging to the second part and their sum is sumsecond.

The difference in sums of the two parts is denoted as sumfirst - sumsecond.

For example, if sumfirst = 3 and sumsecond = 2, their difference is 1. Similarly, if sumfirst = 2 and sumsecond = 3, their difference is -1.

Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,1,4,2,5,6]

› Output: -1

💡 Note: Remove [4,2,3] to get [1,5,6]. First part: [1], Second part: [5,6]. Difference: 1 - (5+6) = -10. This is one possible arrangement; optimal gives -1.

Example 2 — All Same Values

$ Input: nums = [7,7,7,7,7,7]

› Output: 0

💡 Note: All elements equal, so any removal gives same sums for both parts. Difference is always 0.

Example 3 — Small Array

$ Input: nums = [20,40,20,70,30,50]

› Output: -30

💡 Note: Remove [70,40,50] to get [20,20,30]. First: [20], Second: [20,30]. Difference: 20-50 = -30.

Constraints

nums.length == 3 * n
1 ≤ n ≤ 10⁵
1 ≤ nums[i] ≤ 10⁵

Visualization

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Understanding the Visualization

Input Array

Start with 3n elements that need to be processed

Remove n Elements

Strategically remove n elements to optimize the remaining parts

Calculate Difference

Divide remaining 2n elements and find sumfirst - sumsecond

Key Takeaway

🎯 Key Insight: To minimize difference, make first part sum as small as possible and second part sum as large as possible by strategically removing middle elements

Asked in

G Google 15 a Amazon 8 M Microsoft 6

The key insight is to minimize the first part sum and maximize the second part sum. Use heaps to efficiently track the best n elements for each part at every possible split position. Best approach uses heap-based optimization with Time: O(n log n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Heap-Based Optimal Solution	O(n log n)	O(n)	Use heaps to efficiently track best elements for first and second parts
Brute Force - Try All Combinations	O(C(3n,n) × n)	O(n)	Generate all possible ways to remove n elements and calculate minimum difference

Heap-Based Optimal Solution — Algorithm Steps

Pre-compute prefix sums of smallest n elements from left using min-heap
Pre-compute suffix sums of largest n elements from right using max-heap
For each valid split position, calculate minimum difference

Visualization

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Step-by-Step Walkthrough

Prefix Pass

Use max-heap to maintain smallest n elements from left

Suffix Pass

Use min-heap to maintain largest n elements from right

Combine Results

For each split position, calculate minimum difference

Code -

solution.c — C

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

// Simple implementation without full heap - using sorting approach
int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize) {
    int n = numsSize / 3;
    
    // Simplified approach: for each possible split, sort and select optimal elements
    long long minDiff = LLONG_MAX;
    
    for (int split = 0; split <= n; split++) {
        // Left part: select smallest n elements from first n+split elements
        int leftArr[2000];
        for (int i = 0; i < n + split; i++) {
            leftArr[i] = nums[i];
        }
        qsort(leftArr, n + split, sizeof(int), compare);
        
        long long sumFirst = 0;
        for (int i = 0; i < n; i++) {
            sumFirst += leftArr[i];
        }
        
        // Right part: select largest n elements from last n+n-split elements
        int rightArr[2000];
        int rightSize = 0;
        for (int i = n + split; i < numsSize; i++) {
            rightArr[rightSize++] = nums[i];
        }
        qsort(rightArr, rightSize, sizeof(int), compare);
        
        long long sumSecond = 0;
        for (int i = rightSize - n; i < rightSize; i++) {
            sumSecond += rightArr[i];
        }
        
        long long diff = sumFirst - sumSecond;
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return (int)minDiff;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array - simplified parsing
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Two passes with heap operations, each element added/removed from heap takes O(log n)

⚡ Linearithmic

Space Complexity

O(n)

Space for heaps and prefix/suffix arrays

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Heap-Based Optimal Solution — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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