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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Understanding the Visualization

Input

Array [2,-1,1,-2] with change value k=3

Find Pivots

Check each position where left sum could equal right sum

Output

Maximum number of valid partitions after changing at most one element

Key Takeaway

🎯 Key Insight: Use difference tracking to efficiently count how many pivots each element change can balance

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

Approach	Time	Space	Notes
✓ Prefix Sum with Hash Map Optimization	O(n²)	O(n)	Use prefix sums and hash maps to efficiently track balance changes
Single Pass with Difference Tracking	O(n)	O(n)	Track differences and use hash maps to count optimal partitions in one pass
Brute Force - Try All Combinations	O(n³)	O(1)	Check every pivot with every possible element change

Prefix Sum with Hash Map Optimization — Algorithm Steps

Calculate prefix sums for O(1) range sum queries
For each pivot, compute left-right difference
Use hash map to track how changing each element affects differences
Count maximum valid pivots efficiently

Visualization

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Step-by-Step Walkthrough

Build Prefix

Pre-calculate cumulative sums

Quick Sums

Use prefix[pivot] and total-prefix[pivot] for O(1) sums

Count Valid

Check balance for each element change efficiently

Code -

solution.c — C

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

int solution(int* nums, int n, int k) {
    if (n == 1) return 0;
    
    // Calculate prefix sums
    long long* prefix = (long long*)calloc(n + 1, sizeof(long long));
    for (int i = 0; i < n; i++) {
        prefix[i + 1] = prefix[i] + nums[i];
    }
    
    long long totalSum = prefix[n];
    int maxWays = 0;
    
    // Try no change first
    int count = 0;
    for (int pivot = 1; pivot < n; pivot++) {
        long long leftSum = prefix[pivot];
        long long rightSum = totalSum - leftSum;
        if (leftSum == rightSum) count++;
    }
    if (count > maxWays) maxWays = count;
    
    // Try changing each element
    for (int i = 0; i < n; i++) {
        long long changeDiff = k - nums[i];
        count = 0;
        
        for (int pivot = 1; pivot < n; pivot++) {
            long long leftSum = prefix[pivot];
            long long rightSum = totalSum - leftSum;
            
            // Adjust sums if change affects this side
            if (i < pivot) {
                leftSum += changeDiff;
            } else {
                rightSum += changeDiff;
            }
            
            if (leftSum == rightSum) count++;
        }
        
        if (count > maxWays) maxWays = count;
    }
    
    free(prefix);
    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(n²)

Two nested loops: outer for each element change, inner for each pivot position

⚠ Quadratic Growth

Space Complexity

O(n)

Hash map stores up to n different difference values

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Prefix Sum with Hash Map Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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