Introduction

In this blog post, we will explore how to solve the LeetCode problem "1013" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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

Explanation

Here's a breakdown of the solution:

• Calculate the total sum: First, calculate the sum of all elements in the array. If the total sum is not divisible by 3, it's impossible to partition the array into three equal parts, so return False.
• Iterate and find partitions: Iterate through the array, keeping track of the current sum. Whenever the current sum equals total_sum / 3, increment a counter representing the number of partitions found.

• Check for three partitions: Return True if we find at least three partitions (meaning we've found two split points that create three equal-sum subarrays).

• Runtime Complexity: O(n), where n is the length of the array.

• Storage Complexity: O(1).

Code

    def canThreePartsEqualSum(arr):
    total_sum = sum(arr)
    if total_sum % 3 != 0:
        return False

    target_sum = total_sum / 3
    current_sum = 0
    partition_count = 0

    for i in range(len(arr)):
        current_sum += arr[i]
        if current_sum == target_sum:
            partition_count += 1
            current_sum = 0

    return partition_count >= 3