Given an integer array of N elements. The task is to find the number of ways to split the array into two equal sum sub-arrays of non-zero lengths.
Examples:
Input: arr[] = {0, 0, 0, 0}
Output: 3
Explanation:
All the possible ways are: { {{0}, {0, 0, 0}}, {{0, 0}, {0, 0}}, {{0, 0, 0}, {0}}
Therefore, the required output is 3.Input: {1, -1, 1, -1}
Output: 1
Simple Solution: A simple solution is to generate all the possible contiguous sub-arrays pairs and find there sum. If their sum is the same, we will increase the count by one.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The idea is to take an auxiliary array say, aux[] to calculate the presum of the array such that for an index i aux[i] will store the sum of all elements from index 0 to index i.
By doing this we can calculate the left sum and right sum for every index of the array in constant time.
So, the idea is to:
- Find the sum of all the numbers of the array and store it in a variable say, S. If the sum is odd, then the answer will be 0.
- Traverse the array and keep calculating the sum of elements. At ith step, we will use the variable S to maintain sum of all the elements from index 0 to i.
- Calculate sum upto the ith index.
- If this sum equals S/2, increase the count of the number of ways by 1.
- Do this from i=0 to i=N-2.
Below is the implementation of the above approach:
C++
// C++ program to count the number of ways to// divide an array into two halves// with the same sum#include <bits/stdc++.h>using namespace std;// Function to count the number of ways to// divide an array into two halves// with same sumint cntWays(int arr[], int n){ // if length of array is 1 // answer will be 0 as we have // to split it into two // non-empty halves if (n == 1) return 0; // variables to store total sum, // current sum and count int tot_sum = 0, sum = 0, ans = 0; // finding total sum for (int i = 0; i < n; i++) tot_sum += arr[i]; // checking if sum equals total_sum/2 for (int i = 0; i < n - 1; i++) { sum += arr[i]; if (sum == tot_sum / 2) ans++; } return ans;}// Driver Codeint main(){ int arr[] = { 1, -1, 1, -1, 1, -1 }; int n = sizeof(arr) / sizeof(int); cout << cntWays(arr, n); return 0;} |
Java
// Java program to count the number of ways to// divide an array into two halves// with the same sumimport java.util.*;import java.io.*;class GFG { // Function to count the number of ways to // divide an array into two halves // with same sum static int cntWays(int arr[], int n) { // if length of array is 1 // answer will be 0 as we have // to split it into two // non-empty halves if (n == 1) { return 0; } // variables to store total sum, // current sum and count int tot_sum = 0, sum = 0, ans = 0; // finding total sum for (int i = 0; i < n; i++) { tot_sum += arr[i]; } // checking if sum equals total_sum/2 for (int i = 0; i < n - 1; i++) { sum += arr[i]; if (sum == tot_sum / 2) { ans++; } } return ans; } // Driver Code public static void main(String[] args) { int arr[] = {1, -1, 1, -1, 1, -1}; int n = arr.length; System.out.println(cntWays(arr, n)); }}// This code contributed by Rajput-Ji |
Python3
# Python program to count the number of ways to # divide an array into two halves # with the same sum # Function to count the number of ways to # divide an array into two halves # with same sum def cntWays(arr, n): # if length of array is 1 # answer will be 0 as we have # to split it into two # non-empty halves if (n == 1): return 0; # variables to store total sum, # current sum and count tot_sum = 0; sum = 0; ans = 0; # finding total sum for i in range(0,n): tot_sum += arr[i]; # checking if sum equals total_sum/2 for i in range(0,n-1): sum += arr[i]; if (sum == tot_sum / 2): ans+=1; return ans; # Driver Code arr = [1, -1, 1, -1, 1, -1 ]; n = len(arr); print(cntWays(arr, n)); # This code contributed by PrinciRaj1992 |
C#
// C# program to count the number of ways to// divide an array into two halves with// the same sumusing System; class GFG { // Function to count the number of ways to // divide an array into two halves // with same sum static int cntWays(int []arr, int n) { // if length of array is 1 // answer will be 0 as we have // to split it into two // non-empty halves if (n == 1) { return 0; } // variables to store total sum, // current sum and count int tot_sum = 0, sum = 0, ans = 0; // finding total sum for (int i = 0; i < n; i++) { tot_sum += arr[i]; } // checking if sum equals total_sum/2 for (int i = 0; i < n - 1; i++) { sum += arr[i]; if (sum == tot_sum / 2) { ans++; } } return ans; } // Driver Code public static void Main() { int []arr = {1, -1, 1, -1, 1, -1}; int n = arr.Length; Console.WriteLine(cntWays(arr, n)); }}// This code contributed by anuj_67.. |
Javascript
<script> // JavaScript program to count the number of ways to // divide an array into two halves // with the same sum // Function to count the number of ways to // divide an array into two halves // with same sum function cntWays(arr, n) { // if length of array is 1 // answer will be 0 as we have // to split it into two // non-empty halves if (n == 1) return 0; // variables to store total sum, // current sum and count var tot_sum = 0, sum = 0, ans = 0; // finding total sum for (var i = 0; i < n; i++) tot_sum += arr[i]; // checking if sum equals total_sum/2 for (var i = 0; i < n - 1; i++) { sum += arr[i]; if (sum == tot_sum / 2) ans++; } return ans; } // Driver Code var arr = [1, -1, 1, -1, 1, -1]; var n = arr.length; document.write(cntWays(arr, n)); </script> |
Output:
2
Time Complexity: O(N), since there runs a loop from 0 to (n – 1)
Auxiliary Space: O(1), since no extra space has been taken.
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