Generate an array B[] from the given array A[] which satisfies the given conditions

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 even number of odd integers so that the sum can be 0. So for a valid input, there will always be an answer.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to print
// the array elements
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
  
// Function to generate and print
// the required array
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) {
  
            // 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 code
int main()
{
    int arr[] = { 3, -5, -7, 9, 2, -2 };
    int n = sizeof(arr) / sizeof(int);
  
    generateArr(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
// Utility function to print
// the array elements
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] + " ");
}
  
// Function to generate and print
// the required array
static 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 code
public 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

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Python3

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# Python3 implementation of the approach
from math import ceil, floor
  
# Utility function to print
# the array elements
def printArr(arr, n):
    for i in range(n):
        print(arr[i], end = " ")
  
# Function to generate and print
# the required array
def 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 prthe 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 code
arr = [3, -5, -7, 9, 2, -2]
n = len(arr)
  
generateArr(arr, n)
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
// Utility function to print
// the array elements
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] + " ");
}
  
// Function to generate and print
// the required array
static 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 code
public 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

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Output:

1 -2 -4 5 1 -1


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