Given an array A[] of N integers such that A[0] + A[1] + A[2] + … A[N – 1] = 0. The task is to generate an array B[] such that B[i] is either ?A[i] / 2? or ?A[i] / 2? for all valid i and B[0] + B[1] + B[2] + … + B[N – 1] = 0.
Examples:
Input: A[] = {1, 2, -5, 3, -1}
Output: 0 1 -2 1 0
Input: A[] = {3, -5, -7, 9, 2, -2}
Output: 1 -2 -4 5 1 -1
Approach: For even integers, it is safe to assume that B[i] will be A[i] / 2 but for odd integers, to maintain the sum equal to zero, take the ceil of exactly half of odd integers and floor of exactly other half odd integers. Since Odd – Odd = Even and Even – Even = Even and 0 is also Even, it can be said that A[] will always contain an even number of odd integers so that the sum can be 0. So for valid input, there will always be an answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Utility function to print// the array elementsvoid printArr(int arr[], int n)
{ for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}// Function to generate and print// the required arrayvoid generateArr(int arr[], int n)
{ // To switch the ceil and floor
// function alternatively
bool flip = true;
// For every element of the array
for (int i = 0; i < n; i++) {
// If the number is odd then print the ceil
// or floor value after division by 2
if (arr[i] & 1) {
// Use the ceil and floor alternatively
if (flip ^= true)
cout << ceil((float)(arr[i]) / 2.0) << " ";
else
cout << floor((float)(arr[i]) / 2.0) << " ";
}
// If arr[i] is even then it will
// be completely divisible by 2
else {
cout << arr[i] / 2 << " ";
}
}
}// Driver codeint main()
{ int arr[] = { 3, -5, -7, 9, 2, -2 };
int n = sizeof(arr) / sizeof(int);
generateArr(arr, n);
return 0;
} |
Java
// Java implementation of the approach// Utility function to print// the array elementsimport java.util.*;
import java.lang.*;
class GFG
{static void printArr(int arr[], int n)
{ for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}// Function to generate and print// the required arraystatic void generateArr(int arr[], int n)
{ // To switch the ceil and floor
// function alternatively
boolean flip = true;
// For every element of the array
for (int i = 0; i < n; i++)
{
// If the number is odd then print the ceil
// or floor value after division by 2
if ((arr[i] & 1) != 0)
{
// Use the ceil and floor alternatively
if (flip ^= true)
System.out.print((int)(Math.ceil(arr[i] /
2.0)) + " ");
else
System.out.print((int)(Math.floor(arr[i] /
2.0)) + " ");
}
// If arr[i] is even then it will
// be completely divisible by 2
else
{
System.out.print(arr[i] / 2 +" ");
}
}
}// Driver codepublic static void main(String []args)
{ int arr[] = { 3, -5, -7, 9, 2, -2 };
int n = arr.length;
generateArr(arr, n);
}}// This code is contributed by Surendra_Gangwar |
Python3
# Python3 implementation of the approachfrom math import ceil, floor
# Utility function to print# the array elementsdef printArr(arr, n):
for i in range(n):
print(arr[i], end = " ")
# Function to generate and print# the required arraydef generateArr(arr, n):
# To switch the ceil and floor
# function alternatively
flip = True
# For every element of the array
for i in range(n):
# If the number is odd then print the ceil
# or floor value after division by 2
if (arr[i] & 1):
# Use the ceil and floor alternatively
flip ^= True
if (flip):
print(int(ceil((arr[i]) / 2)),
end = " ")
else:
print(int(floor((arr[i]) / 2)),
end = " ")
# If arr[i] is even then it will
# be completely divisible by 2
else:
print(int(arr[i] / 2), end = " ")
# Driver codearr = [3, -5, -7, 9, 2, -2]
n = len(arr)
generateArr(arr, n)# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach// Utility function to print// the array elementsusing System;
using System.Collections.Generic;
class GFG
{static void printArr(int []arr, int n)
{ for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}// Function to generate and print// the required arraystatic void generateArr(int []arr, int n)
{ // To switch the ceil and floor
// function alternatively
bool flip = true;
// For every element of the array
for (int i = 0; i < n; i++)
{
// If the number is odd then print the ceil
// or floor value after division by 2
if ((arr[i] & 1) != 0)
{
// Use the ceil and floor alternatively
if (flip ^= true)
Console.Write((int)(Math.Ceiling(arr[i] /
2.0)) + " ");
else
Console.Write((int)(Math.Floor(arr[i] /
2.0)) + " ");
}
// If arr[i] is even then it will
// be completely divisible by 2
else
{
Console.Write(arr[i] / 2 +" ");
}
}
}// Driver codepublic static void Main(String []args)
{ int []arr = { 3, -5, -7, 9, 2, -2 };
int n = arr.Length;
generateArr(arr, n);
}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript implementation of the approach// Utility function to print// the array elements function printArr(arr , n) {
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Function to generate and print
// the required array
function generateArr(arr , n) {
// To switch the ceil and floor
// function alternatively
var flip = true;
// For every element of the array
for (i = 0; i < n; i++) {
// If the number is odd then print the ceil
// or floor value after division by 2
if ((arr[i] & 1) != 0) {
// Use the ceil and floor alternatively
if (flip ^= true)
document.write(parseInt( (Math.ceil(arr[i] / 2.0))) + " ");
else
document.write(parseInt( (Math.floor(arr[i] / 2.0))) + " ");
}
// If arr[i] is even then it will
// be completely divisible by 2
else {
document.write(arr[i] / 2 + " ");
}
}
}
// Driver code
var arr = [ 3, -5, -7, 9, 2, -2 ];
var n = arr.length;
generateArr(arr, n);
// This code is contributed by todaysgaurav </script> |
Output:
1 -2 -4 5 1 -1
Time Complexity: O(n)
Auxiliary Space: O(1)
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