Count Partitions with Even Sum Difference

Count Partitions with Even Sum Difference - Problem

You are given an integer array nums of length n. A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that:

Left subarray contains indices [0, i]
Right subarray contains indices [i + 1, n - 1]

Return the number of partitions where the difference between the sum of the left and right subarrays is even.

Input & Output

Example 1 — Basic Case

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

› Output: 3

💡 Note: Partition at i=0: left=[1], right=[2,3,4], difference=|1-9|=8 (even). Partition at i=1: left=[1,2], right=[3,4], difference=|3-7|=4 (even). Partition at i=2: left=[1,2,3], right=[4], difference=|6-4|=2 (even). All 3 partitions have even differences, so count = 3.

Example 2 — Different Result

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

› Output: 2

💡 Note: Partition at i=0: left=[1], right=[2,3], difference=|1-5|=4 (even). Partition at i=1: left=[1,2], right=[3], difference=|3-3|=0 (even). Both partitions have even differences, so count = 2.

Example 3 — All Odd Differences

$ Input: nums = [1,2]

› Output: 0

💡 Note: Only one partition at i=0: left=[1], right=[2], difference=|1-2|=1 (odd). No partitions with even difference, so count = 0.

Constraints

2 ≤ nums.length ≤ 10⁵
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

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

The key insight is recognizing that a difference is even when both left and right sums have the same parity (both odd or both even). Best approach uses mathematical property with running sum: Time O(n), Space O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each possible partition position i, calculate the sum of left subarray [0, i] and right subarray [i+1, n-1], then check if their difference is even.

Mathematical Insight

⏱️ Time: O(n) Space: O(1)

Key insight: leftSum - rightSum is even if and only if leftSum and rightSum have the same parity (both even or both odd). Since rightSum = total - leftSum, this happens when total and 2×leftSum have same parity.

Prefix Sum

⏱️ Time: O(n) Space: O(n)

Precompute prefix sums to get any subarray sum in constant time. For each partition, left sum is prefix[i] and right sum is total - prefix[i].

Brute Force — Algorithm Steps

Iterate through each partition position i from 0 to n-2
Calculate sum of left subarray [0, i]
Calculate sum of right subarray [i+1, n-1]
Check if |leftSum - rightSum| is even

Visualization

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Step-by-Step Walkthrough

Position 0

Calculate left=[1], right=[2,3,4] sums

Position 1

Calculate left=[1,2], right=[3,4] sums

Position 2

Calculate left=[1,2,3], right=[4] sums

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int count = 0;
    
    for (int i = 0; i < n - 1; i++) {
        int leftSum = 0;
        for (int j = 0; j <= i; j++) {
            leftSum += nums[j];
        }
        
        int rightSum = 0;
        for (int j = i + 1; j < n; j++) {
            rightSum += nums[j];
        }
        
        if ((leftSum - rightSum) % 2 == 0) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000];
    int n = 0;
    
    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        nums[n++] = (int)strtol(p, &p, 10);
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n-1 positions, calculate sums taking O(n) time

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses variables for sums and counter

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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