Given an array arr[] of N integers with total element sum equal to zero. The task is to reduce every element to it’s half such that the total sum remain zero. For every odd element X in the array, it could be reduced to either(X + 1) / 2 or (X – 1) / 2.
Examples:
Input: arr[] = {-7, 14, -7}
Output: -4 7 -3
-4 + 7 -3 = 0
Input: arr[] = {-14, 14}
Output: -7 7
Approach: All the even elements could be divided by 2 but for odd elements, they have to be alternatively reduced to (X + 1) / 2 and (X – 1) / 2 in order to retain the original sum (i.e. 0) in the final array.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include<bits/stdc++.h>using namespace std;// Function to reduce every // element to it's half such that // the total sum remain zerovoid half(int arr[], int n){ int i; // Flag to switch between alternating // odd numbers in the array int flag = 0; // For every element of the array for (i = 0; i < n; i++) { // If its even then reduce it to half if (arr[i] % 2 == 0 ) cout << arr[i] / 2 << " "; // If its odd else { // Reduce the odd elements // alternatively if (flag == 0) { cout << arr[i] / 2 - 1 << " "; // Switch flag flag = 1; } else { int q = arr[i] / 2; cout<<q <<" "; // Switch flag flag = 0; } } }}// Driver code int main () { int arr[] = {-7, 14, -7}; int len = sizeof(arr)/sizeof(arr[0]); half(arr, len) ; return 0;}// This code is contributed by Rajput-Ji |
Java
// Java implementation of the above approach class GFG {// Function to reduce every // element to it's half such that // the total sum remain zero static void half(int arr[], int n){ int i; // Flag to switch between alternating // odd numbers in the array int flag = 0; // For every element of the array for (i = 0; i < n; i++) { // If its even then reduce it to half if (arr[i] % 2 == 0 ) System.out.print(arr[i] / 2 + " "); // If its odd else { // Reduce the odd elements // alternatively if (flag == 0) { System.out.print(arr[i] / 2 - 1 + " "); // Switch flag flag = 1; } else { int q = arr[i] / 2; System.out.print(q + " "); // Switch flag flag = 0; } } }}// Driver code public static void main (String[] args) { int arr[] = {-7, 14, -7}; int len = arr.length; half(arr, len) ;}}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to reduce every# element to it's half such that # the total sum remain zerodef half(arr, n) : # Flag to switch between alternating # odd numbers in the array flag = 0 # For every element of the array for i in range(n): # If its even then reduce it to half if arr[i] % 2 == 0 : print(arr[i]//2, end =" ") # If its odd else : # Reduce the odd elements # alternatively if flag == 0: print(arr[i]//2, end =" ") # Switch flag flag = 1 else : q = arr[i]//2 q+= 1 print(q, end =" ") # Switch flag flag = 0# Driver codearr = [-7, 14, -7]half(arr, len(arr)) |
C#
// C# implementation of the above approach using System;class GFG {// Function to reduce every // element to it's half such that // the total sum remain zero static void half(int []arr, int n){ int i; // Flag to switch between alternating // odd numbers in the array int flag = 0; // For every element of the array for (i = 0; i < n; i++) { // If its even then reduce it to half if (arr[i] % 2 == 0 ) Console.Write(arr[i] / 2 + " "); // If its odd else { // Reduce the odd elements // alternatively if (flag == 0) { Console.Write(arr[i] / 2 - 1 + " "); // Switch flag flag = 1; } else { int q = arr[i] / 2; Console.Write(q + " "); // Switch flag flag = 0; } } }}// Driver code public static void Main () { int [] arr = {-7, 14, -7}; int len = arr.Length; half(arr, len) ;}}// This code is contributed by mohit kumar 29 |
Javascript
<script>// Javascript implementation of the above approach// Function to reduce every// element to it's half such that// the total sum remain zerofunction half(arr, n){ let i; // Flag to switch between alternating // odd numbers in the array let flag = 0; // For every element of the array for (i = 0; i < n; i++) { // If its even then reduce it to half if (arr[i] % 2 == 0 ) document.write(arr[i] / 2 + " "); // If its odd else { // Reduce the odd elements // alternatively if (flag == 0) { document.write(Math.ceil(arr[i] / 2) - 1 + " "); // Switch flag flag = 1; } else { let q = Math.ceil(arr[i] / 2); document.write(q + " "); // Switch flag flag = 0; } } }}// Driver code let arr = [-7, 14, -7]; let len = arr.length; half(arr, len) ;// This code is contributed by _saurabh_jaiswal</script> |
Output:
-4 7 -3
Time Complexity : O(n) ,as we are traversing once on the array.
Space Complexity : O(1) ,as we are not using any extra space.
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