Open in App
Not now

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

• Difficulty Level : Hard
• Last Updated : 06 Sep, 2022

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```

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 ` `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

 `= (``\$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

 ``

Output

```Yes
```

Complexity Analysis:

Time Complexity : O(n+m), since there runs two loops, one for n times and the other for m times.
Auxiliary Complexity: O(m), since m extra space has been taken.

My Personal Notes arrow_drop_up
Related Articles