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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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

✓ Brute Force - Try All Combinations

⏱️ Time: O(C(3n,n) × n) Space: O(n)

Try every possible combination of n elements to remove, then for each remaining 2*n elements, calculate sumfirst - sumsecond and track the minimum difference.

Heap-Based Optimal Solution

⏱️ Time: O(n log n) Space: O(n)

For each possible split position, use min-heap to find smallest n elements for first part from left side, and max-heap to find largest n elements for second part from right side.

Brute Force - Try All Combinations — Algorithm Steps

Generate all C(3n, n) combinations of n elements to remove
For each combination, calculate sums of first n and second n remaining elements
Track minimum difference across all combinations

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all C(3n,n) ways to remove n elements

Calculate Differences

For each combination, compute sumfirst - sumsecond

Find Minimum

Track the smallest difference found

Code -

solution.c — C

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

static int combinations[100000][20];
static int combCount;
static int remaining[20];
static int nums[20];

void generateCombinations(int start, int total, int k, int* current, int currentSize) {
    if (k == 0) {
        for (int i = 0; i < currentSize; i++) {
            combinations[combCount][i] = current[i];
        }
        combinations[combCount][currentSize] = -1; // marker
        combCount++;
        return;
    }
    for (int i = start; i <= total - k; i++) {
        current[currentSize] = i;
        generateCombinations(i + 1, total, k - 1, current, currentSize + 1);
    }
}

int solution(int* numsArr, int numsSize) {
    int n = numsSize / 3;
    int minDiff = INT_MAX;
    
    // Copy to global array for easier access
    for (int i = 0; i < numsSize; i++) {
        nums[i] = numsArr[i];
    }
    
    // Generate all combinations of n indices to remove
    combCount = 0;
    int current[20];
    generateCombinations(0, numsSize, n, current, 0);
    int removeCombCount = combCount;
    
    for (int c = 0; c < removeCombCount; c++) {
        // Create remove set
        int removeSet[20] = {0};
        for (int i = 0; combinations[c][i] != -1; i++) {
            removeSet[combinations[c][i]] = 1;
        }
        
        // Build remaining array
        int remainingSize = 0;
        for (int i = 0; i < numsSize; i++) {
            if (!removeSet[i]) {
                remaining[remainingSize++] = nums[i];
            }
        }
        
        // Split remaining 2n elements: first n vs next n
        int sumFirst = 0, sumSecond = 0;
        for (int i = 0; i < n; i++) {
            sumFirst += remaining[i];
        }
        for (int i = n; i < remainingSize; i++) {
            sumSecond += remaining[i];
        }
        
        int diff = sumFirst - sumSecond;
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return minDiff;
}

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;
        
        char* endptr;
        long val = strtol(p, &endptr, 10);
        if (endptr != p) {
            arr[(*size)++] = (int)val;
            p = endptr;
        } else {
            p++;
        }
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int numsArr[20];
    int numsSize;
    parseArray(line, numsArr, &numsSize);
    
    int result = solution(numsArr, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(C(3n,n) × n)

C(3n,n) combinations to try, each requiring O(n) time to calculate sums

✓ Linear Growth

Space Complexity

O(n)

Storage for current combination and remaining elements

⚡ Linearithmic Space

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Medium Frequency

~35 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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