LeetCode #1013 — EASY

Partition Array Into Three Parts With Equal Sum

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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])

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

Constraints:

3 <= arr.length <= 5 * 10⁴
-10⁴ <= arr[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: 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])

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[0,2,1,-6,6,-7,9,1,2,0,1]

Example 2

[0,2,1,-6,6,7,9,-1,2,0,1]

Example 3

[3,3,6,5,-2,2,5,1,-9,4]

Core Insight

What unlocks the optimal approach

If we have three parts with the same sum, what is the sum of each? If you can find the first part, can you find the second part?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1013: Partition Array Into Three Parts With Equal Sum
class Solution {
    public boolean canThreePartsEqualSum(int[] arr) {
        int s = Arrays.stream(arr).sum();
        if (s % 3 != 0) {
            return false;
        }
        s /= 3;
        int cnt = 0, t = 0;
        for (int x : arr) {
            t += x;
            if (t == s) {
                cnt++;
                t = 0;
            }
        }
        return cnt >= 3;
    }
}

// Accepted solution for LeetCode #1013: Partition Array Into Three Parts With Equal Sum
func canThreePartsEqualSum(arr []int) bool {
	s := 0
	for _, x := range arr {
		s += x
	}
	if s%3 != 0 {
		return false
	}
	s /= 3
	cnt, t := 0, 0
	for _, x := range arr {
		t += x
		if t == s {
			cnt++
			t = 0
		}
	}
	return cnt >= 3
}

# Accepted solution for LeetCode #1013: Partition Array Into Three Parts With Equal Sum
class Solution:
    def canThreePartsEqualSum(self, arr: List[int]) -> bool:
        s, mod = divmod(sum(arr), 3)
        if mod:
            return False
        cnt = t = 0
        for x in arr:
            t += x
            if t == s:
                cnt += 1
                t = 0
        return cnt >= 3

// Accepted solution for LeetCode #1013: Partition Array Into Three Parts With Equal Sum
impl Solution {
    pub fn can_three_parts_equal_sum(arr: Vec<i32>) -> bool {
        let sum: i32 = arr.iter().sum();
        let s = sum / 3;
        let mod_val = sum % 3;
        if mod_val != 0 {
            return false;
        }

        let mut cnt = 0;
        let mut t = 0;
        for &x in &arr {
            t += x;
            if t == s {
                cnt += 1;
                t = 0;
            }
        }

        cnt >= 3
    }
}

// Accepted solution for LeetCode #1013: Partition Array Into Three Parts With Equal Sum
function canThreePartsEqualSum(arr: number[]): boolean {
    let s = arr.reduce((a, b) => a + b);
    if (s % 3) {
        return false;
    }
    s = (s / 3) | 0;
    let [cnt, t] = [0, 0];
    for (const x of arr) {
        t += x;
        if (t == s) {
            cnt++;
            t = 0;
        }
    }
    return cnt >= 3;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.