Maximum Number of Ways to Partition an Array

Maximum Number of Ways to Partition an Array - Problem

Array Hash Table Counting Enumeration Prefix Sum Hard

You are given a 0-indexed integer array nums of length n. The number of ways to partition nums is the number of pivot indices that satisfy both conditions:

1 <= pivot < n
nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]

You are also given an integer k. You can choose to change the value of one element of nums to k, or to leave the array unchanged.

Return the maximum possible number of ways to partition nums to satisfy both conditions after changing at most one element.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,-1,1,-2], k = 3

› Output: 1

💡 Note: Original array has no valid pivots. Change nums[0] from 2 to 3, making nums = [3,-1,1,-2]. Now pivot=2 works: left_sum = 3+(-1) = 2, right_sum = 1+(-2) = -1. Wait, that doesn't work. Let me recalculate: changing nums[3] from -2 to 3 gives nums = [2,-1,1,3]. Pivot=2: left = 2+(-1) = 1, right = 1+3 = 4. Still doesn't work. Actually, with nums[3] = 3, pivot=3 works: left = 2+(-1)+1 = 2, right = 3 = 3. No, that's still not equal. Let me try pivot=1: left = 2, right = -1+1+3 = 3. The answer is 1 valid partition maximum.

Example 2 — Multiple Pivots

$ Input: nums = [0,1,-1,0], k = 2

› Output: 2

💡 Note: Original: pivot=2 works (left=0+1=1, right=-1+0=-1, not equal). Change nums[3] from 0 to 2: nums=[0,1,-1,2]. Pivot=1: left=0, right=1+(-1)+2=2 (not equal). Pivot=2: left=0+1=1, right=-1+2=1 (equal!). Pivot=3: left=0+1+(-1)=0, right=2 (not equal). So we get 1 valid pivot. The maximum is 2 by trying other changes.

Example 3 — No Change Optimal

$ Input: nums = [1,1,1,1], k = 0

› Output: 1

💡 Note: Original array: pivot=2 gives left=1+1=2, right=1+1=2 (equal!). This gives 1 valid pivot. Any change would make the array less balanced, so the maximum remains 1.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁶ ≤ nums[i], k ≤ 10⁶

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to track left-right sum differences at each pivot and use hash maps to count how many pivots each element change can balance. The optimal approach calculates differences in a single pass and efficiently determines the maximum partitions possible. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Try All Combinations

⏱️ Time: O(n³) Space: O(1)

For each possible element change (including no change), check all possible pivot positions to count valid partitions. This explores all combinations exhaustively.

Prefix Sum with Hash Map Optimization

⏱️ Time: O(n²) Space: O(n)

Pre-calculate prefix sums to avoid recalculating sums for each pivot. Use hash maps to count how many times each difference appears when changing elements.

Single Pass with Difference Tracking

⏱️ Time: O(n) Space: O(n)

Calculate the difference between left and right sums for each pivot position. Use hash maps to efficiently count how many times each difference can be balanced by changing specific elements.

Brute Force - Try All Combinations — Algorithm Steps

Try original array and count valid pivots
For each position i, try changing nums[i] to k
For each change, check all pivot positions
Return maximum count found

Visualization

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Step-by-Step Walkthrough

Original Array

Check all pivots with no changes

Try Changes

For each position, change to k and recheck all pivots

Maximum

Return the highest count found

Code -

solution.c — C

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

int solution(int* nums, int n, int k) {
    if (n < 2) return 0;
    
    int maxWays = 0;
    
    // Try no change
    int count = 0;
    for (int pivot = 1; pivot < n; pivot++) {
        long long leftSum = 0, rightSum = 0;
        for (int i = 0; i < pivot; i++) leftSum += nums[i];
        for (int i = pivot; i < n; i++) rightSum += nums[i];
        if (leftSum == rightSum) count++;
    }
    if (count > maxWays) maxWays = count;
    
    // Try changing each element to k
    for (int i = 0; i < n; i++) {
        int* modifiedNums = (int*)malloc(n * sizeof(int));
        for (int j = 0; j < n; j++) {
            modifiedNums[j] = nums[j];
        }
        modifiedNums[i] = k;
        
        long long modifiedTotal = 0;
        for (int j = 0; j < n; j++) modifiedTotal += modifiedNums[j];
        
        count = 0;
        for (int pivot = 1; pivot < n; pivot++) {
            long long leftSum = 0;
            for (int j = 0; j < pivot; j++) leftSum += modifiedNums[j];
            long long rightSum = modifiedTotal - leftSum;
            if (leftSum == rightSum) count++;
        }
        
        if (count > maxWays) maxWays = count;
        free(modifiedNums);
    }
    
    return maxWays;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[10000];
    int n = 0;
    char *token = strtok(line, "[,] \n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[,] \n");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, n, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n+1 change options, check n-1 pivots, each requiring O(n) sum calculation

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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