Last Updated: 5 Mar, 2021
Split Array With Equal Sums
Moderate
Asked in companies
You are given an array/list 'ARR' of size 'N'. You task is to find if there exists a triplet (i, j, k) such that 0 < i , i + 1 < j , j + 1 < k and k < 'N' - 1 and the sum of the subarrays [0, i - 1],[i + 1, j - 1], [j + 1, k - 1], [k + 1, N - 1] are equal.
An array c is a subarray of array d if c can be obtained from d by deletion of several elements from the beginning and several elements from the end.
Example:
let 'ARR' = [1, 2, 3] then the possible subarrays of 'ARR' will be - {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 2, 3}.
Note: Assume That the Array has Zero-based indexing.
Input format:
The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows.
The first line of each test case contains an integer ‘N’ representing the array’s size.
The second line of each test case contains 'N' space-separated integers representing the array’s elements.
Output format:
For each test case, print ‘True’ if such a triplet exists; else print ‘False’.
Output for each test case will be printed in a separate line.
Note:
You don’t have to take any input or print anything; it already has been taken care of. Just implement the function.
Constraints:
1 <= T <= 5
1 <= N <= 10 ^ 3
-10 ^ 6 <= ARR[i] <= 10 ^ 6
Time Limit: 1 sec
Approaches
To implement this approach, You have to understand the constraints first, i.e., 0 < i, i+1 < j , j+1< k < N-1.
If you carefully look at the constraints, you may notice the problem while splitting the array into slices or subarrays does not include the elements at index i, j, and k. So it can be simply observed that according to constraints during the slices you make, three of the elements will not get included in any of the slices. Every Slice needs to be non-empty, too, so your task is to make four of these partitions. You can conclude from these points that the array size should be at least 7 to find such a triplet, any size below 7 will result in ‘False’.
To simplify this we can write this as:
0 < i < N - 5
i + 1 < j < N - 3
j + 1 < k < N - 1
The algorithm will be:
- We can iterate over each ‘i’ from 1 to ‘N’ - 5:
- Now for each ‘j’ from ‘i’ + 1 to ‘N’ - 3:
- Now for each 'k' from 'j' + 1 to 'N' - 1:
- We can calculate the sum for each of the above-formed subarray in one iteration.
- If the sum of the subarrays is equal we return ‘True’.
- Now for each 'k' from 'j' + 1 to 'N' - 1:
- Now for each ‘j’ from ‘i’ + 1 to ‘N’ - 3:
- If we do not find any valid triplet we return ‘False’.
This one is the optimization to the Brute force approach. We will iterate over all possible ‘i’ , ‘j’ and ‘k’ and for each possible subarray instead of calculating it’s sum in O(N) time complexity we can get in O(1) time complexity using prefix sums and pre calculation.
You can do this by first dividing the arraylist in two equal sums and then further dividing the first half in two equal sums and then checking the remaining half for these sums.
- Initially store the prefix sum of the array in a separate array.
- For all the possible j
- Check the value for which the sum of the right half and the left half is equal.
- Now for all possible values of i in the left half.
- Check for the equal half sums, and store the sum in a hash map.
- Now for each possible k in the right half.
- Check for the equal half sums, and store the sum in a hash map.
- If any common value is found in both of the hashmaps we return ‘True’.
- If no valid triplet is found we return ‘False’.