Find the Value of the Partition - Problem

Array Sorting Medium

You are given a positive integer array nums.

Partition nums into two arrays, nums1 and nums2, such that:

Each element of the array nums belongs to either the array nums1 or the array nums2.
Both arrays are non-empty.
The value of the partition is minimized.

The value of the partition is |max(nums1) - min(nums2)|.

Here, max(nums1) denotes the maximum element of the array nums1, and min(nums2) denotes the minimum element of the array nums2.

Return the integer denoting the value of such partition.

Input & Output

Example 1 — Basic Case

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

› Output: 1

💡 Note: Sort to get [1,2,3,4]. Best partition is [1] and [2,3,4] giving |1-2| = 1, or [1,2] and [3,4] giving |2-3| = 1.

Example 2 — All Same Elements

$ Input: nums = [100,1,10]

› Output: 9

💡 Note: Sort to get [1,10,100]. Check splits: |1-10| = 9, |10-100| = 90. Minimum is 9.

Example 3 — Two Elements

$ Input: nums = [1,1]

› Output: 0

💡 Note: Only one way to partition: nums1=[1], nums2=[1]. Value is |1-1| = 0.

Constraints

2 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 A Amazon 8

The key insight is that the optimal partition always occurs between consecutive elements in the sorted array. Sort the array and check all adjacent pairs to find the minimum difference. Best approach is Sort and Check: Time: O(n log n), Space: O(1).

Common Approaches

✓ Brute Force - Try All Partitions

⏱️ Time: O(2^n) Space: O(n)

Generate all possible ways to split the array into two non-empty groups. For each partition, calculate |max(nums1) - min(nums2)| and track the minimum value.

Sort and Check Consecutive Pairs

⏱️ Time: O(n log n) Space: O(1)

Sort the array first, then check all possible partitions where we split between consecutive elements. The optimal partition always involves consecutive elements in the sorted array.

Brute Force - Try All Partitions — Algorithm Steps

Generate all 2^n - 2 possible partitions
For each partition, find max of first group and min of second group
Calculate absolute difference and track minimum

Visualization

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Step-by-Step Walkthrough

Generate Partitions

Try all 2^n ways to split array

Calculate Values

For each partition, find |max(nums1) - min(nums2)|

Find Minimum

Return the smallest partition value

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int minValue = INT_MAX;
    
    // Try all possible partitions
    for (int mask = 1; mask < (1 << n) - 1; mask++) {
        int nums1[1000], nums2[1000];
        int len1 = 0, len2 = 0;
        
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) {
                nums1[len1++] = nums[i];
            } else {
                nums2[len2++] = nums[i];
            }
        }
        
        if (len1 > 0 && len2 > 0) {
            int maxNums1 = nums1[0];
            for (int i = 1; i < len1; i++) {
                if (nums1[i] > maxNums1) maxNums1 = nums1[i];
            }
            
            int minNums2 = nums2[0];
            for (int i = 1; i < len2; i++) {
                if (nums2[i] < minNums2) minNums2 = nums2[i];
            }
            
            int value = abs(maxNums1 - minNums2);
            if (value < minValue) minValue = value;
        }
    }
    
    return minValue;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Generate 2^n partitions, each taking O(n) time to evaluate

⚡ Linearithmic

Space Complexity

O(n)

Space for recursion stack and storing current partition

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Find the Value of the Partition - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Partitions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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