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Check if at least half array is reducible to zero by performing some operations

  • Difficulty Level : Hard
  • Last Updated : 30 Apr, 2021

Given an array of n-positive elements. In each operation, you can select some elements and decrease them by 1 and increase remaining elements by m. The task is to determine that whether after some iterations is it possible to have at least half of elements of given array equal to zero or not.
Examples
 

Input : arr[] = {3, 5, 6, 8}, m = 2 
Output : Yes

Input : arr[] = {4, 7, 12, 13, 34},  m = 7
Output : No

 

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If we try to analyze the problem statement we will find that at any step either you are decreasing an element by 1 or increasing it by m. That means on every single step performed, there are total three possibilities if we compare two elements.
Let a1 and a2 are two elements then: 
 



  1. Either both element got decreased by 1 and hence resulting no change in their actual difference.
  2. Both element got increased by m and hence resulting no change in their actual difference again.
  3. One element got decreased by 1 and other one got increased by m and hence resulting a change of (m+1) in their actual difference.

It means on every step you are going to keep the difference between any two element either same or increase it by (m+1).
So if you take the modulo of all element by (m+1) and keep its frequency we can check how many elements can be made equal to zero at any point of time.
Algorithm :
 

  • Create a hash table of size m+1
  • Capture the frequency of elements as (arr[i] % (m+1)) and store in hash table.
  • Find out the maximum frequency and if it is greater than or equals to n/2, then the answer is YES otherwise NO.

Below is the implementation of the above approach: 
 

C++




// C++ program to find whether half-array
// reducible to 0
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the desired
// result after computation
void isHalfReducible(int arr[], int n, int m)
{
    int frequencyHash[m + 1];
    int i;
 
    memset(frequencyHash, 0, sizeof(frequencyHash));
    for (i = 0; i < n; i++) {
        frequencyHash[arr[i] % (m + 1)]++;
    }
 
    for (i = 0; i <= m; i++) {
        if (frequencyHash[i] >= n / 2)
            break;
    }
 
    if (i <= m)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 16, 32, 3, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int m = 7;
 
    isHalfReducible(arr, n, m);
 
    return 0;
}

Java




// Java Program to find whether half-array
// reducible to 0
 
public class GFG {
 
    // Function to print the desired
    // result after computation
    static void isHalfReducible(int arr[], int n, int m)
    {
        int frequencyHash[] = new int[m + 1];
        int i;
 
        for(i = 0 ; i < frequencyHash.length ; i++)
            frequencyHash[i] = 0 ;
        
        for (i = 0; i < n; i++) {
            frequencyHash[arr[i] % (m + 1)]++;
        }
 
        for (i = 0; i <= m; i++) {
            if (frequencyHash[i] >= n / 2)
                break;
        }
 
        if (i <= m)
            System.out.println("Yes") ;
        else
            System.out.println("No") ;
    }
 
// Driver code
    public static void main(String args[])
    {
            int arr[] = { 8, 16, 32, 3, 12 };
            int n = arr.length ;
 
            int m = 7;
 
            isHalfReducible(arr, n, m);
 
    }
    // This code is contributed by ANKITRAI1
}

Python3




# Python3 program to find whether
# half-array reducible to 0
 
# Function to print the desired
# result after computation
def isHalfReducible(arr, n, m):
     
    frequencyHash =[0]*(m + 1);
    i = 0;
    while(i < n):
        frequencyHash[(arr[i] % (m + 1))] += 1;
        i += 1;
 
    i = 0;
    while(i <= m):
        if(frequencyHash[i] >= (n / 2)):
            break;
        i += 1;
 
    if (i <= m):
        print("Yes");
    else:
        print("No");
 
# Driver Code
arr = [ 8, 16, 32, 3, 12 ];
n = len(arr);
 
m = 7;
isHalfReducible(arr, n, m);
     
# This code is contributed by mits

C#




// C# Program to find whether half-array
// reducible to 0
 
using System;
 
  
public class GFG {
  
    // Function to print the desired
    // result after computation
    static void isHalfReducible(int[] arr, int n, int m)
    {
        int[] frequencyHash = new int[m + 1];
        int i;
  
        for(i = 0 ; i < frequencyHash.Length ; i++)
            frequencyHash[i] = 0 ;
         
        for (i = 0; i < n; i++) {
            frequencyHash[arr[i] % (m + 1)]++;
        }
  
        for (i = 0; i <= m; i++) {
            if (frequencyHash[i] >= n / 2)
                break;
        }
  
        if (i <= m)
            Console.WriteLine("Yes") ;
        else
            Console.WriteLine("No") ;
    }
  
// Driver code
    public static void Main()
    {
            int[] arr = { 8, 16, 32, 3, 12 };
            int n = arr.Length ;
  
            int m = 7;
  
            isHalfReducible(arr, n, m);
  
    }
    // This code is contributed by Subhadeep
}

PHP




<?php
// PHP program to find whether 
// half-array reducible to 0
 
// Function to print the desired
// result after computation
function isHalfReducible($arr, $n, $m)
{
    $frequencyHash = array_fill(0, $m + 1, 0);
    $i = 0;
    for (;$i < $n; $i++)
    {
        $frequencyHash[($arr[$i] % ($m + 1))]++;
    }
 
    for ($i = 0; $i <= $m; $i++)
    {
        if ($frequencyHash[$i] >= ($n / 2))
            break;
    }
 
    if ($i <= $m)
        echo "Yes\n";
    else
        echo "No\n";
}
 
// Driver Code
$arr = array( 8, 16, 32, 3, 12 );
$n = sizeof($arr);
 
$m = 7;
isHalfReducible($arr, $n, $m);
     
// This code is contributed by mits
?>

Javascript




<script>
 
// javascript program to find whether half-array
// reducible to 0
 
// Function to print the desired
// result after computation
function isHalfReducible(arr, n, m)
{
    var frequencyHash = Array(m+1).fill(0);
    var i;
 
    for (i = 0; i < n; i++) {
        frequencyHash[arr[i] % (m + 1)]++;
    }
 
    for (i = 0; i <= m; i++) {
        if (frequencyHash[i] >= n / 2)
            break;
    }
 
    if (i <= m)
        document.write( "Yes" );
    else
        document.write( "No" );
}
 
// Driver Code
var arr = [ 8, 16, 32, 3, 12 ];
var n = arr.length;
var m = 7;
isHalfReducible(arr, n, m);
 
</script>
Output: 
Yes

 




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