Array Partition

Array Partition - Problem

Given an integer array nums of 2n integers, group these integers into n pairs (a₁, b₁), (a₂, b₂), ..., (aₙ, bₙ) such that the sum of min(aᵢ, bᵢ) for all i is maximized.

Return the maximized sum.

Input & Output

Example 1 — Basic Case

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

› Output: 4

💡 Note: After sorting: [1,2,3,4]. Pairs: (1,2) and (3,4). Sum of minimums: min(1,2) + min(3,4) = 1 + 3 = 4

Example 2 — Larger Array

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

› Output: 9

💡 Note: After sorting: [1,2,2,5,6,6]. Pairs: (1,2), (2,5), (6,6). Sum of minimums: 1 + 2 + 6 = 9

Example 3 — Minimum Size

$ Input: nums = [1,2]

› Output: 1

💡 Note: Only one pair possible: (1,2). Sum of minimums: min(1,2) = 1

Constraints

1 ≤ n ≤ 10⁴
nums.length == 2 * n
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

a Amazon 25 M Microsoft 18 f Facebook 15 G Google 12

The key insight is that to maximize the sum of minimums, we should pair each element with the next larger element. Best approach is to sort the array and sum elements at even indices. Time: O(n log n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Pairings

⏱️ Time: O(n!) Space: O(n)

Try every possible way to group the array into pairs and calculate the sum of minimum values for each grouping. Return the maximum sum found.

Sort First - Greedy Approach

⏱️ Time: O(n log n) Space: O(1)

Sort the array in ascending order, then pair adjacent elements. This ensures that each minimum contributes optimally to the sum.

Brute Force - Try All Pairings — Algorithm Steps

Generate all possible pairings of the array
For each pairing, calculate sum of minimums
Return the maximum sum

Visualization

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Step-by-Step Walkthrough

Generate Pairings

Create all possible ways to pair elements

Calculate Sums

For each pairing, sum the minimum of each pair

Find Maximum

Return the highest sum found

Code -

solution.c — C

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

int maxSum = 0;

void generatePairings(int* arr, int arrSize, int pairs[][2], int pairCount) {
    if (arrSize == 0) {
        int currentSum = 0;
        for (int i = 0; i < pairCount; i++) {
            currentSum += (pairs[i][0] < pairs[i][1]) ? pairs[i][0] : pairs[i][1];
        }
        maxSum = (currentSum > maxSum) ? currentSum : maxSum;
        return;
    }
    
    int first = arr[0];
    for (int i = 1; i < arrSize; i++) {
        pairs[pairCount][0] = first;
        pairs[pairCount][1] = arr[i];
        
        int remaining[20];
        int remainingSize = 0;
        for (int j = 1; j < arrSize; j++) {
            if (j != i) {
                remaining[remainingSize++] = arr[j];
            }
        }
        
        generatePairings(remaining, remainingSize, pairs, pairCount + 1);
    }
}

int solution(int* nums, int numsSize) {
    maxSum = 0;
    int pairs[10][2];
    generatePairings(nums, numsSize, pairs, 0);
    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);
    
    int nums[20];
    int numsSize = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums, numsSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n!)

Generating all possible pairings requires factorial time

⚡ Linearithmic

Space Complexity

O(n)

Space for storing current pairing arrangement

✓ Linear Space

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Input & Output

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Related Problems

Common Approaches

Brute Force - Try All Pairings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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