Problem Statement

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 + 1 < j with (arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])

LeetCode

Constraints

3 <= arr.length <= 5 * 10^4
-10^4 <= arr[i] <= 10^4

Examples

Example 1:

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

Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1

Example 2:

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

Example 3:

Input: arr = [3,3,6,5,-2,2,5,1,-9,4]
Output: true

Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4

Different Approaches

1️⃣ Brute Force Approach

Code:

class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& arr) {
        int n = arr.size();

        // Step 1: Calculate the total sum of the array
        int totalSum = 0;
        for (int num : arr) {
            totalSum += num;
        }

        // Step 2: If total sum is not divisible by 3, we cannot split it into 3 equal parts
        if (totalSum % 3 != 0) {
            return false;
        }

        int targetSum = totalSum / 3;

        // Step 3: Try all possible ways to split the array into 3 parts
        // First cut can be at index i (left part: 0 to i)
        for (int i = 0; i < n - 2; i++) {
            int leftSum = 0;

            // Calculate the sum from index 0 to i (first part)
            for (int k = 0; k <= i; k++) {
                leftSum += arr[k];
            }

            // If the first part doesn't sum to the target, skip
            if (leftSum != targetSum) {
                continue;
            }

            // Second cut can be at index j (middle part: i+1 to j)
            for (int j = i + 1; j < n - 1; j++) {
                int middleSum = 0;

                // Calculate the sum from index i+1 to j (second part)
                for (int k = i + 1; k <= j; k++) {
                    middleSum += arr[k];
                }

                // If the second part doesn't sum to the target, skip
                if (middleSum != targetSum) {
                    continue;
                }

                // Step 4: The remaining part (j+1 to end) must also sum to the target
                int rightSum = 0;
                for (int k = j + 1; k < n; k++) {
                    rightSum += arr[k];
                }

                // If all three parts match the target sum, return true
                if (rightSum == targetSum) {
                    return true;
                }
            }
        }

        // If no valid split found, return false
        return false;
    }
};

Complexity Analysis:

• Time Complexity:O(n^3)
• Space Complexity:O(1)

2️⃣ Prefix Sum Approach (Optimal Approach)

Approach:

1. Compute totalSum.
   If totalSum % 3 != 0 → return false.
2. Target sum for each part = totalSum / 3.
3. Traverse array, keep currSum.
   Count how many times currSum == target.
4. If you find 2 such partitions (i.e., 2 cuts), and you're not at the end, third part is valid.

🧠 Why This Works:

1. Total sum divisible by 3 → each part must sum to targetSum.
2. You scan from left to right, collecting sums.
3. Every time you reach a targetSum, you reset the sum and increment the part count.
4. Once you've found 2 valid parts, the rest of the array automatically becomes the 3rd part.

Code:

class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& arr) {
        // Step 1: Calculate the total sum of the array
        int totalSum = accumulate(arr.begin(), arr.end(), 0);

        // Step 2: If total sum is not divisible by 3, we can't split it equally into 3 parts
        if (totalSum % 3 != 0) return false;

        int targetSum = totalSum / 3;  // Each part must sum to this value

        int currentSum = 0;
        int partsFound = 0;

        // Step 3: Traverse the array and count how many times we hit the target sum
        for (int i = 0; i < arr.size(); ++i) {
            currentSum += arr[i];

            // Whenever we reach the target sum, we count it as one valid part
            if (currentSum == targetSum) {
                partsFound++;       // Found one part
                currentSum = 0;     // Reset to start tracking the next part

                // Step 4: If we already found 2 parts, and we're not at the end,
                // the remaining elements automatically make the third part.
                if (partsFound == 2 && i < arr.size() - 1) {
                    return true;
                }
            }
        }

        // Step 5: If we never found two valid parts, return false
        return false;
    }
};

Complexity Analysis:

• Time Complexity: O(n)
• Space Complexity: O(1)