Partition Array Into Three Parts With Equal Sum

Partition Array Into Three Parts With Equal Sum - Problem

Array Greedy Easy

Given an array of integers arr, return true if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i and j with (i + 1 < j) such that:

arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1]

Input & Output

Example 1 — Equal Partition Possible

$ Input: arr = [0,2,1,-6,6,7,9,-1,2,0,1]

› Output: true

💡 Note: We can partition into [0,2,1,-6,6], [7], [9,-1,2,0,1] where each part has sum = 5. Actually, let me use a clearer example.

Example 2 — Simple Case

$ Input: arr = [3,3,3]

› Output: true

💡 Note: We can partition into [3], [3], [3] where each part has sum = 3

Example 3 — Impossible Partition

$ Input: arr = [0,2,1,-6,6,18]

› Output: false

💡 Note: Total sum is 21, target per part is 7. But we cannot find valid partition points that create three parts with sum 7 each

Constraints

3 ≤ arr.length ≤ 5 × 10⁴
-10⁴ ≤ arr[i] ≤ 10⁴

Visualization

Tap to expand

Asked in

A Amazon 15 G Google 8 F Facebook 6

The key insight is that total sum must be divisible by 3 for equal partition to exist. Best approach uses greedy single pass to find partition points efficiently. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Check every valid combination of two partition points i and j where i+1 < j, and verify if all three parts have equal sums.

Greedy Single Pass

⏱️ Time: O(n) Space: O(1)

First check if the total sum is divisible by 3 (necessary condition). If yes, scan the array once to find the first partition point where sum equals total/3, then find the second partition point.

Prefix Sum Optimization

⏱️ Time: O(n²) Space: O(n)

Precompute prefix sums to avoid recalculating sums for each partition attempt. This reduces the time complexity from O(n³) to O(n²).

Brute Force — Algorithm Steps

Step 1: Try all valid pairs (i,j) where i+1 < j
Step 2: Calculate sum of each of the three parts
Step 3: Check if all three sums are equal

Visualization

Tap to expand

Step-by-Step Walkthrough

Try i=0, j=2

Check if parts [0], [1], [2,3,4] have equal sums

Try i=0, j=3

Check if parts [0], [1,2], [3,4] have equal sums

Continue

Test all valid (i,j) pairs until equal partition found

Code -

solution.c — C

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

bool solution(int* arr, int n) {
    // Try all valid partition points i and j
    for (int i = 0; i < n - 2; i++) {
        for (int j = i + 2; j < n; j++) {
            // Calculate sum of first part [0...i]
            int sum1 = 0;
            for (int k = 0; k <= i; k++) sum1 += arr[k];
            
            // Calculate sum of second part [i+1...j-1]
            int sum2 = 0;
            for (int k = i + 1; k < j; k++) sum2 += arr[k];
            
            // Calculate sum of third part [j...n-1]
            int sum3 = 0;
            for (int k = j; k < n; k++) sum3 += arr[k];
            
            // Check if all three sums are equal
            if (sum1 == sum2 && sum2 == sum3) {
                return true;
            }
        }
    }
    
    return false;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Nested loops for i and j positions, plus O(n) to calculate sums for each pair

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

32.0K Views

Medium Frequency

~15 min Avg. Time

890 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 — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler