Count of elements such that difference between sum of left and right sub arrays is equal to a multiple of k
Given an array arr[] of length n and an integer k, the task is to find the number of indices from 2 to n-1 in the array having a difference of the sum of left and right sub array equal to the multiple of given number k.
Examples:
Input: arr[] = {1, 2, 3, 3, 1, 1}, k = 4
Output: 2
Explanation: The only possible indices are 4 and 5Input: arr[] = {1, 2, 3, 4, 5}, k = 1
Output: 3
Approach:
- Create a prefix array which contains the sum of the elements in the left and the suffix array which contains the sum of elements in the right.
- Check for every index the difference of sum in the left and right and increase the counter if it is divisible by k
Below is the implementation of the above approach:
CPP
// C++ code to count of elements such that // difference between the sum of left and right // sub-arrays are equal to a multiple of k #include <bits/stdc++.h> using namespace std; // Functions to find the no of elements int noOfElement(int a[], int n, int k) { // Creating a prefix array int prefix[n]; // Starting element of prefix array // will be the first element // of given array prefix[0] = a[0]; for (int i = 1; i < n; i++) { prefix[i] = prefix[i - 1] + a[i]; } // Creating a suffix array; int suffix[n]; // Last element of suffix array will // be the last element of given array suffix[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { suffix[i] = suffix[i + 1] + a[i]; } // Checking difference of left and right half // is divisible by k or not. int cnt = 0; for (int i = 1; i < n - 1; i++) { if ((prefix[i] - suffix[i]) % k == 0 || (suffix[i] - prefix[i]) % k == 0) { cnt = cnt + 1; } } return cnt; } // Driver code int main() { int a[] = { 1, 2, 3, 3, 1, 1 }; int k = 4; int n = sizeof(a) / sizeof(a[0]); cout << noOfElement(a, n, k); return 0; } |
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Java
// Java code to count of elements such that // difference between the sum of left and right // sub-arrays are equal to a multiple of k class GFG { // Functions to find the no of elements static int noOfElement(int a[], int n, int k) { // Creating a prefix array int []prefix = new int[n]; // Starting element of prefix array // will be the first element // of given array prefix[0] = a[0]; for (int i = 1; i < n; i++) { prefix[i] = prefix[i - 1] + a[i]; } // Creating a suffix array; int []suffix = new int[n]; // Last element of suffix array will // be the last element of given array suffix[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { suffix[i] = suffix[i + 1] + a[i]; } // Checking difference of left and right half // is divisible by k or not. int cnt = 0; for (int i = 1; i < n - 1; i++) { if ((prefix[i] - suffix[i]) % k == 0 || (suffix[i] - prefix[i]) % k == 0) { cnt = cnt + 1; } } return cnt; } // Driver code public static void main(String[] args) { int a[] = { 1, 2, 3, 3, 1, 1 }; int k = 4; int n = a.length; System.out.print(noOfElement(a, n, k)); } } // This code is contributed by Rajput-Ji |
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Python
# Python3 code to count of elements such that # difference between the sum of left and right # sub-arrays are equal to a multiple of k # Functions to find the no of elements def noOfElement(a, n, k): # Creating a prefix array prefix = [0] * n # Starting element of prefix array # will be the first element # of given array prefix[0] = a[0] for i in range(1, n): prefix[i] = prefix[i - 1] + a[i] # Creating a suffix array suffix = [0] * n # Last element of suffix array will # be the last element of given array suffix[n - 1] = a[n - 1] for i in range(n - 2, -1, -1): suffix[i] = suffix[i + 1] + a[i] # Checking difference of left and right half # is divisible by k or not. cnt = 0 for i in range(1, n - 1): if ((prefix[i] - suffix[i]) % k == 0 or (suffix[i] - prefix[i]) % k == 0): cnt = cnt + 1 return cnt # Driver code a = [ 1, 2, 3, 3, 1, 1 ] k = 4n = len(a) print(noOfElement(a, n, k)) # This code is contributed by mohit kumar 29 |
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C#
// C# code to count of elements such that // difference between the sum of left and right // sub-arrays are equal to a multiple of k using System; class GFG { // Functions to find the no of elements static int noOfElement(int []a, int n, int k) { // Creating a prefix array int []prefix = new int[n]; // Starting element of prefix array // will be the first element // of given array prefix[0] = a[0]; for (int i = 1; i < n; i++) { prefix[i] = prefix[i - 1] + a[i]; } // Creating a suffix array; int []suffix = new int[n]; // Last element of suffix array will // be the last element of given array suffix[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) { suffix[i] = suffix[i + 1] + a[i]; } // Checking difference of left and right half // is divisible by k or not. int cnt = 0; for (int i = 1; i < n - 1; i++) { if ((prefix[i] - suffix[i]) % k == 0 || (suffix[i] - prefix[i]) % k == 0) { cnt = cnt + 1; } } return cnt; } // Driver code public static void Main() { int []a = { 1, 2, 3, 3, 1, 1 }; int k = 4; int n = a.Length; Console.Write(noOfElement(a, n, k)); } } // This code is contributed by AnkitRai01 |
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Output:
2
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