Given an array of positive integers of length n. Our task is to find minimum number of operations to convert an array so that arr[i] % 4 is zero for each i. In each operation, we can take any two elements from the array, remove both of them and put back their sum in the array.
Examples:
Input : arr = {2 , 2 , 2 , 3 , 3}
Output : 3
Explanation: In 1 operation we pick 2 and 2 and put their sum back to the array , In 2 operation we pick 3 and 3 and do same for that ,now in 3 operation we pick 6 and 2 so overall 3 operation are required.Input: arr = {4, 2, 2, 6, 6}
Output: 2
Explanation: In operation 1, we can take 2 and 2 and put back their sum i.e. 4. In operation 2, we can take 6 and 6 and put back their sum i.e. 12. And array becomes {4, 4, 12}.
Approach : Assume the count of elements leaving remainder 1, 2, 3 when divided by 4 are brr[1], brr[2] and brr[3].
If (brr[1] + 2 * brr[2] + 3 * brr[3]) is not a multiple of 4, solution does not exist.
Now greedily pair elements of brr[2] with brr[2] and elements of brr[1] with brr[3]. This helps us to achieve fixing a maximum of 2 elements at a time. Now, we can either we left with only 1 brr[2] element or none. If we are left with 1 brr[2] element, then we can pair with 2 remaining brr[1] or brr[3] elements. This will incur a total of 2 operations.
At last, we would be only left with brr[1] or brr[3] elements (if possible). This can only we fixed in one way. That is taking 4 of them and fixing them all together in 3 operations. Thus, we are able to fix all the elements of the array.
Below is the implementation:
C++
// CPP program to find Minimum number// of operations to convert an array // so that arr[i] % 4 is zero.#include <bits/stdc++.h>using namespace std;// Function to find minimum operations.int minimumOperations(int arr[], int n){ // Counting of all the elements // leaving remainder 1, 2, 3 when // divided by 4 in the array brr. // at positions 1, 2 and 3 respectively. int brr[] = { 0, 0, 0, 0 }; for (int i = 0; i < n; i++) brr[arr[i] % 4]++; // If it is possible to convert the // array so that arr[i] % 4 is zero. if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0) { // Pairing the elements of brr3 and brr1. int min_opr = min(brr[3], brr[1]); brr[3] -= min_opr; brr[1] -= min_opr; // Pairing the brr2 elements. min_opr += brr[2] / 2; // Assigning the remaining brr2 elements. brr[2] %= 2; // If we are left with one brr2 element. if (brr[2]) { // Here we need only two operations // to convert the remaining one // brr2 element to convert it. min_opr += 2; // Now there is no brr2 element. brr[2] = 0; // Remaining brr3 elements. if (brr[3]) brr[3] -= 2; // Remaining brr1 elements. if (brr[1]) brr[1] -= 2; } // If we are left with brr1 and brr2 // elements then, we have to take four // of them and fixing them all together // in 3 operations. if (brr[1]) min_opr += (brr[1] / 4) * 3; if (brr[3]) min_opr += (brr[3] / 4) * 3; // Returns the minimum operations. return min_opr; } // If it is not possible to convert the array. return -1; }// Driver functionint main(){ int arr[] = { 1, 2, 3, 1, 2, 3, 8 }; int n = sizeof(arr) / sizeof(arr[0]); cout << minimumOperations(arr, n);} |
Java
// Java program to find Minimum number// of operations to convert an array // so that arr[i] % 4 is zero.class GFG { // Function to find minimum operations.static int minimumOperations(int arr[], int n){ // Counting of all the elements // leaving remainder 1, 2, 3 when // divided by 4 in the array brr. // at positions 1, 2 and 3 respectively. int brr[] = { 0, 0, 0, 0 }; for (int i = 0; i < n; i++) brr[arr[i] % 4]++; // If it is possible to convert the // array so that arr[i] % 4 is zero. if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0) { // Pairing the elements of brr3 and brr1. int min_opr = Math.min(brr[3], brr[1]); brr[3] -= min_opr; brr[1] -= min_opr; // Pairing the brr2 elements. min_opr += brr[2] / 2; // Assigning the remaining brr2 elements. brr[2] %= 2; // If we are left with one brr2 element. if (brr[2] == 1) { // Here we need only two operations // to convert the remaining one // brr2 element to convert it. min_opr += 2; // Now there is no brr2 element. brr[2] = 0; // Remaining brr3 elements. if (brr[3] == 1) brr[3] -= 2; // Remaining brr1 elements. if (brr[1]== 1) brr[1] -= 2; } // If we are left with brr1 and brr2 // elements then, we have to take four // of them and fixing them all together // in 3 operations. if (brr[1] != 0) min_opr += (brr[1] / 4) * 3; if (brr[3] != 0) min_opr += (brr[3] / 4) * 3; // Returns the minimum operations. return min_opr; } // If it is not possible to convert the array. return -1; } // Driver functionpublic static void main(String[] args){ int arr[] = { 1, 2, 3, 1, 2, 3, 8 }; int n = arr.length; System.out.println(minimumOperations(arr, n));}}// This code is contributed by Prerna Saini. |
Python3
# Python program to# find Minimum number# of operations to# convert an array # so that arr[i] % 4 is zero.# Function to find# minimum operations.def minimumOperations(arr,n): # Counting of all the elements # leaving remainder 1, 2, 3 when # divided by 4 in the array brr. # at positions 1, 2 and 3 respectively. brr = [ 0, 0, 0, 0 ] for i in range(n): brr[arr[i] % 4]+=1; # If it is possible to convert the # array so that arr[i] % 4 is zero. if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0): # Pairing the elements # of brr3 and brr1. min_opr = min(brr[3], brr[1]) brr[3] -= min_opr brr[1] -= min_opr # Pairing the brr2 elements. min_opr += brr[2] // 2 # Assigning the remaining # brr2 elements. brr[2] %= 2 # If we are left with # one brr2 element. if (brr[2]): # Here we need only two operations # to convert the remaining one # brr2 element to convert it. min_opr += 2 # Now there is no brr2 element. brr[2] = 0 # Remaining brr3 elements. if (brr[3]): brr[3] -= 2 # Remaining brr1 elements. if (brr[1]): brr[1] -= 2 # If we are left with brr1 and brr2 # elements then, we have to take four # of them and fixing them all together # in 3 operations. if (brr[1]): min_opr += (brr[1] // 4) * 3 if (brr[3]): min_opr += (brr[3] // 4) * 3 # Returns the minimum operations. return min_opr # If it is not possible to convert the array. return -1 # Driver functionarr = [ 1, 2, 3, 1, 2, 3, 8 ]n =len(arr)print(minimumOperations(arr, n))# This code is contributed# by Anant Agarwal. |
C#
// C# program to find Minimum number// of operations to convert an array // so that arr[i] % 4 is zero.using System;class GFG { // Function to find minimum operations. static int minimumOperations(int []arr, int n) { // Counting of all the elements // leaving remainder 1, 2, 3 when // divided by 4 in the array brr. // at positions 1, 2 and 3 respectively. int []brr = { 0, 0, 0, 0 }; for (int i = 0; i < n; i++) brr[arr[i] % 4]++; // If it is possible to convert the // array so that arr[i] % 4 is zero. if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0) { // Pairing the elements of brr3 and brr1. int min_opr = Math.Min(brr[3], brr[1]); brr[3] -= min_opr; brr[1] -= min_opr; // Pairing the brr2 elements. min_opr += brr[2] / 2; // Assigning the remaining brr2 elements. brr[2] %= 2; // If we are left with one brr2 element. if (brr[2] == 1) { // Here we need only two operations // to convert the remaining one // brr2 element to convert it. min_opr += 2; // Now there is no brr2 element. brr[2] = 0; // Remaining brr3 elements. if (brr[3] == 1) brr[3] -= 2; // Remaining brr1 elements. if (brr[1]== 1) brr[1] -= 2; } // If we are left with brr1 and brr2 // elements then, we have to take four // of them and fixing them all together // in 3 operations. if (brr[1] == 1) min_opr += (brr[1] / 4) * 3; if (brr[3] == 1) min_opr += (brr[3] / 4) * 3; // Returns the minimum operations. return min_opr; } // If it is not possible to convert the array. return -1; } // Driver function public static void Main() { int []arr = { 1, 2, 3, 1, 2, 3, 8 }; int n = arr.Length; Console.WriteLine(minimumOperations(arr, n)); }}// This code is contributed by vt_m |
PHP
<?php// PHP program to find // Minimum number of // operations to convert // an array so that // arr[i] % 4 is zero.// Function to find // minimum operations.function minimumOperations($arr, $n){ // Counting of all the // elements leaving remainder // 1, 2, 3 when divided by 4 // in the array brr at positions // 1, 2 and 3 respectively. $brr = array(0, 0, 0, 0); for ($i = 0; $i < $n; $i++) $brr[$arr[$i] % 4]++; // If it is possible to // convert the array so // that arr[i] % 4 is zero. if (($brr[1] + 2 * $brr[2] + 3 * $brr[3]) % 4 == 0) { // Pairing the elements // of brr3 and brr1. $min_opr = min($brr[3], $brr[1]); $brr[3] -= $min_opr; $brr[1] -= $min_opr; // Pairing the // brr2 elements. $min_opr += $brr[2] / 2; // Assigning the remaining // brr2 elements. $brr[2] %= 2; // If we are left with // one brr2 element. if ($brr[2]) { // Here we need only two // operations to convert // the remaining one brr2 // element to convert it. $min_opr += 2; // Now there is no // brr2 element. $brr[2] = 0; // Remaining brr3 elements. if ($brr[3]) $brr[3] -= 2; // Remaining brr1 elements. if ($brr[1]) $brr[1] -= 2; } // If we are left with brr1 // and brr2 elements then, // we have to take four of // them and fixing them all // together in 3 operations. if ($brr[1]) $min_opr += ($brr[1] / 4) * 3; if ($brr[3]) $min_opr += ($brr[3] / 4) * 3; // Returns the // minimum operations. return $min_opr; } // If it is not possible // to convert the array. return -1; }// Driver Code$arr = array(1, 2, 3, 1, 2, 3, 8);$n = count($arr);echo (minimumOperations($arr, $n));// This code is contributed by // Manish Shaw(manishshaw1)?> |
Javascript
<script>// Java Script program to find Minimum number// of operations to convert an array // so that arr[i] % 4 is zero. // Function to find minimum operations.function minimumOperations(arr,n){ // Counting of all the elements // leaving remainder 1, 2, 3 when // divided by 4 in the array brr. // at positions 1, 2 and 3 respectively. let brr = [0, 0, 0, 0 ]; for (let i = 0; i < n; i++) brr[arr[i] % 4]++; // If it is possible to convert the // array so that arr[i] % 4 is zero. if ((brr[1] + 2 * brr[2] + 3 * brr[3]) % 4 == 0) { // Pairing the elements of brr3 and brr1. let min_opr = Math.min(brr[3], brr[1]); brr[3] -= min_opr; brr[1] -= min_opr; // Pairing the brr2 elements. min_opr += brr[2] / 2; // Assigning the remaining brr2 elements. brr[2] %= 2; // If we are left with one brr2 element. if (brr[2] == 1) { // Here we need only two operations // to convert the remaining one // brr2 element to convert it. min_opr += 2; // Now there is no brr2 element. brr[2] = 0; // Remaining brr3 elements. if (brr[3] == 1) brr[3] -= 2; // Remaining brr1 elements. if (brr[1]== 1) brr[1] -= 2; } // If we are left with brr1 and brr2 // elements then, we have to take four // of them and fixing them all together // in 3 operations. if (brr[1] == 1) min_opr += (brr[1] / 4) * 3; if (brr[3] == 1) min_opr += (brr[3] / 4) * 3; // Returns the minimum operations. return min_opr; } // If it is not possible to convert the array. return -1; } // Driver function let arr= [1, 2, 3, 1, 2, 3, 8 ]; let n = arr.length; document.write(minimumOperations(arr, n));// This code is contributed by Bobby</script> |
Output
3
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