Split Array With Same Average - Leetcode Solution

Problem Statement

Given an array A of integers, partition it into two non-empty subarrays such that the average value of the elements in the two subarrays is equal. Output "True" if it is possible, otherwise output "False".

Example:

Input: [1,2,3,4,5,6,7,8] Output: True Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both subarrays have the same average.

Solution

Approach #1: Subset Sum Approach

One approach to solve this problem is to use subset sums. First, calculate the sum of all elements in the array. Then, we can try to find a subset of the array whose sum is half of the sum of the whole array. If we find such a subset, it means we can partition the array into two subarrays with the same average.

To find a subset with a given sum, we can use dynamic programming. We can create a boolean table dp[n+1][sum+1], where dp[i][j] is true if there is a subset of elements from 0 to i with a sum of j, and false otherwise. The transition function for the table is:

dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i-1]]

This means that we can either exclude the ith element from the subset or include it. If we include it, we look for a subset with the sum (j-nums[i-1]) in the previous row (i-1).

We start filling the table from dp[0][0], which is true, and fill it row by row. If we find dp[n][sum/2] to be true, it means we can partition the array into two subarrays with the same average.

Time Complexity: O(n*sum), where n is the length of the array and sum is the sum of all elements in the array.

Approach #2: Depth-First Search Approach

Another approach to solve this problem is to use a depth-first search (DFS) with a sliding window technique. We can start with a window of size one and keep increasing the size of the window until we find a subarray with the same average as the whole array.

Let's say the sum of the whole array is s and the length of the whole array is n. If we split the array into two subarrays with the same average, the sum of each subarray would be s/2. Therefore, if we consider a subarray of length k, its sum should be ks/(2n). If we can find such a subarray, we can split the array into two subarrays with the same average.

We can use a sliding window technique to find such a subarray. We fix the starting index of the window and slide it over the array. At each step, we calculate the sum of the subarray and check if it's equal to ks/(2n). If the sum is less than ks/(2n), we can move the window to the right, because adding more elements would increase the sum. If the sum is greater than ks/(2n), we can move the starting index of the window, because removing some elements would decrease the sum. If the sum is equal to ks/(2n), we can increase the size of the window and repeat the process.

We can use DFS to try different values of k (the length of the subarray). We start with k=1 and keep increasing it until we find a subarray with the same average as the whole array. If the size of one subarray is fixed, the size of the other subarray is also fixed, because the sum of both subarrays should be equal to s/2. Therefore, we can calculate the length of the other subarray and check if it's a valid subarray.

Time Complexity: O(2^(n/2)), where n is the length of the array. The worst-case time complexity occurs when we have to check all possible values of k (from 1 to n/2), and the DFS tree has a depth of n/2.

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