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++
#include <bits/stdc++.h>
using namespace std;
void printArr( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
void generateArr( int arr[], int n)
{
bool flip = true ;
for ( int i = 0; i < n; i++) {
if (arr[i] & 1) {
if (flip ^= true )
cout << ceil (( float )(arr[i]) / 2.0) << " " ;
else
cout << floor (( float )(arr[i]) / 2.0) << " " ;
}
else {
cout << arr[i] / 2 << " " ;
}
}
}
int main()
{
int arr[] = { 3, -5, -7, 9, 2, -2 };
int n = sizeof (arr) / sizeof ( int );
generateArr(arr, n);
return 0;
}
|
Java
import 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] + " " );
}
static void generateArr( int arr[], int n)
{
boolean flip = true ;
for ( int i = 0 ; i < n; i++)
{
if ((arr[i] & 1 ) != 0 )
{
if (flip ^= true )
System.out.print(( int )(Math.ceil(arr[i] /
2.0 )) + " " );
else
System.out.print(( int )(Math.floor(arr[i] /
2.0 )) + " " );
}
else
{
System.out.print(arr[i] / 2 + " " );
}
}
}
public static void main(String []args)
{
int arr[] = { 3 , - 5 , - 7 , 9 , 2 , - 2 };
int n = arr.length;
generateArr(arr, n);
}
}
|
Python3
from math import ceil, floor
def printArr(arr, n):
for i in range (n):
print (arr[i], end = " " )
def generateArr(arr, n):
flip = True
for i in range (n):
if (arr[i] & 1 ):
flip ^ = True
if (flip):
print ( int (ceil((arr[i]) / 2 )),
end = " " )
else :
print ( int (floor((arr[i]) / 2 )),
end = " " )
else :
print ( int (arr[i] / 2 ), end = " " )
arr = [ 3 , - 5 , - 7 , 9 , 2 , - 2 ]
n = len (arr)
generateArr(arr, n)
|
C#
using 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] + " " );
}
static void generateArr( int []arr, int n)
{
bool flip = true ;
for ( int i = 0; i < n; i++)
{
if ((arr[i] & 1) != 0)
{
if (flip ^= true )
Console.Write(( int )(Math.Ceiling(arr[i] /
2.0)) + " " );
else
Console.Write(( int )(Math.Floor(arr[i] /
2.0)) + " " );
}
else
{
Console.Write(arr[i] / 2 + " " );
}
}
}
public static void Main(String []args)
{
int []arr = { 3, -5, -7, 9, 2, -2 };
int n = arr.Length;
generateArr(arr, n);
}
}
|
Javascript
<script>
function printArr(arr , n) {
for (i = 0; i < n; i++)
document.write(arr[i] + " " );
}
function generateArr(arr , n) {
var flip = true ;
for (i = 0; i < n; i++) {
if ((arr[i] & 1) != 0) {
if (flip ^= true )
document.write(parseInt( (Math.ceil(arr[i] / 2.0))) + " " );
else
document.write(parseInt( (Math.floor(arr[i] / 2.0))) + " " );
}
else {
document.write(arr[i] / 2 + " " );
}
}
}
var arr = [ 3, -5, -7, 9, 2, -2 ];
var n = arr.length;
generateArr(arr, n);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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