Split Array with Equal Sum

Split Array with Equal Sum - Problem

Given an integer array nums of length n, return true if there is a triplet (i, j, k) which satisfies the following conditions:

0 < i, i + 1 < j, j + 1 < k < n - 1
The sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) is equal.

A subarray (l, r) represents a slice of the original array starting from the element indexed l to the element indexed r.

Note: The array is split into four parts by three dividers at positions i, j, and k.

Input & Output

Example 1 — Valid Split Exists

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

› Output: true

💡 Note: With i=1, j=3, k=5: part1=[1] sum=1, part2=[1] sum=1, part3=[1] sum=1, part4=[1] sum=1. All sums equal 1.

Example 2 — No Valid Split

$ Input: nums = [1,2,3,4,5,6,7]

› Output: false

💡 Note: No triplet (i,j,k) can create four subarrays with equal sums from this ascending sequence.

Example 3 — Too Small Array

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

› Output: false

💡 Note: Array too small (length 4 < 7). Need at least 7 elements for valid triplet positions.

Constraints

7 ≤ nums.length ≤ 2000
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

Tap to expand

Asked in

G Google 25 f Facebook 18 M Microsoft 12

The key insight is to fix the middle divider j and solve two independent matching problems. The hash map approach with prefix sums achieves O(n²) time complexity by avoiding redundant calculations. Best approach: Hash Map with Fixed Middle. Time: O(n²), Space: O(n)

Common Approaches

✓ Brute Force - Try All Triplets

⏱️ Time: O(n⁴) Space: O(1)

For each valid triplet of dividers, calculate the sum of all four subarrays and check if they are equal. This involves three nested loops to try all possible positions.

Hash Map with Fixed Middle

⏱️ Time: O(n²) Space: O(n)

Fix the middle divider j first. For each j, calculate possible sums for left side (i positions) and store in hash map. Then check right side (k positions) to find matches.

Prefix Sum Optimization

⏱️ Time: O(n³) Space: O(n)

Use prefix sums to avoid recalculating subarray sums repeatedly. This reduces the time for sum calculation from O(n) to O(1), improving overall complexity.

Brute Force - Try All Triplets — Algorithm Steps

Iterate through all valid positions for i (1 to n-4)
For each i, iterate through all valid positions for j (i+2 to n-3)
For each j, iterate through all valid positions for k (j+2 to n-2)
Calculate sums of four subarrays and check equality

Visualization

Tap to expand

Step-by-Step Walkthrough

Try i=1

Place first divider at position 1

Try j=3

Place second divider at position 3 (i+2)

Try k=5

Place third divider at position 5 (j+2)

Calculate Sums

Sum each of the four parts and compare

Code -

solution.c — C

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

bool solution(int* nums, int n) {
    if (n < 7) return false;
    
    for (int i = 1; i < n - 5; i++) {
        for (int j = i + 2; j < n - 3; j++) {
            for (int k = j + 2; k < n - 1; k++) {
                int sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
                
                for (int x = 0; x < i; x++) sum1 += nums[x];
                for (int x = i + 1; x < j; x++) sum2 += nums[x];
                for (int x = j + 1; x < k; x++) sum3 += nums[x];
                for (int x = k + 1; x < n; x++) sum4 += nums[x];
                
                if (sum1 == sum2 && sum2 == sum3 && sum3 == sum4) {
                    return true;
                }
            }
        }
    }
    
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000], n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    bool result = solution(nums, n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n⁴)

Three nested loops O(n³) with O(n) sum calculation for each iteration

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

⚡ Linearithmic Space

34.5K Views

Medium Frequency

~25 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Triplets — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler