Given an array Arr[] of N integers and an integer K, the task is to find if it is possible to make the count of even and odd elements equal by performing the following operations at most K times:
- Choose any index i such that Arr[i] is even and divide it by 2.
- Choose any index i such that Arr[i] is odd and multiply it by 2.
Examples:
Input: Arr[] = {1, 4, 8, 12}, K = 2
Output: Yes
Explanation: Count of Odd = 1, Count of Even = 3.
If we half 4 twice then 4 becomes 1 or if we half 12 twice then it becomes 3.
It is possible to make even and odd count equal by performing 2 operations.Input: Arr[] = {1, 2, 3, 4}, K = 0
Output: Yes
Approach: The idea to solve this problem is as follows:
Find the count of even and odd elements (say expressed as CE and CO respectively).
The number of elements needed to be modified = abs(CE – CO)/2.
An even character needed to be halved i times if its right most bit is at (i+1)th position from the right. And an odd element can be made even by multiplying it by 2 in a single operation.Use this criteria to find the number of operations required and if it is at most K or not.
Follow the below steps to solve the problem:
- If N is odd, return False.
- Else Initialize a vector (say v) of size 32 with 0 to store the count of rightmost bits at a position, CE (Count of Even) = 0 and CO(Count of Odd) = 0.
- Iterate through the array from 0 to N-1
- Find CE and CO of Arr[]
- Find the index of the rightmost set bit for every array element and increment that index value of the vector v.
- If CE = CO, then no operations required to make counts equal.
- If CO > CE, the result will be (CO – CE)/2.
- Otherwise, find required operations by iterating the vector.
- Return true If the required operation is less than K. Else return false.
Below is the implementation of the above approach.
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to check if it is possible// to make even and odd element count same// using at most K operationsbool findCount(int* arr, int N, int K){ // Edge Case if (N & 1) { return 0; } // Initialize the variables int Res, CE = 0, CO = 0; vector<int> v(32, 0); // Find the index of rightmost set bit // for every array element and increment // that index of vector, also find even // and odd count for (int i = 0; i < N; i++) { if (arr[i] & 1) CO++; else CE++; v[ffs(arr[i]) - 1]++; } // Condition is already true if (CE == CO) { Res = 0; } // Count of Even > Count of Odd else if (CE > CO) { int n = (CE - CO) / 2; Res = 0; // Find minimum operations to make the // even and odd count equal for (int i = 1; i < v.size(); i++) { if (n >= v[i]) { Res += v[i] * i; n = n - v[i]; } else { Res += n * i; break; } } } // Count of Odd > Count of Even else { Res = (CO - CE) / 2; } return Res <= K;}// Driver Codeint main(){ int Arr[] = { 1, 4, 8, 12 }; int N = sizeof(Arr) / sizeof(Arr[0]); int K = 2; // Function Call if (findCount(Arr, N, K)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java code to implement the above approachimport java.util.*;class GFG{// Function to check if it is possible// to make even and odd element count same// using at most K operationsstatic boolean findCount(int []arr, int N, int K){ // Edge Case if ((N & 1)>0) { return false; } // Initialize the variables int Res, CE = 0, CO = 0; int []v = new int[32]; // Find the index of rightmost set bit // for every array element and increment // that index of vector, also find even // and odd count for (int i = 0; i < N; i++) { if ((arr[i] & 1)>0) CO++; else CE++; v[arr[i] & (arr[i]-1) - 1]++; } // Condition is already true if (CE == CO) { Res = 0; } // Count of Even > Count of Odd else if (CE > CO) { int n = (CE - CO) / 2; Res = 0; // Find minimum operations to make the // even and odd count equal for (int i = 1; i < v.length; i++) { if (n >= v[i]) { Res += v[i] * i; n = n - v[i]; } else { Res += n * i; break; } } } // Count of Odd > Count of Even else { Res = (CO - CE) / 2; } return Res <= K;}// Driver Codepublic static void main(String[] args){ int Arr[] = { 1, 4, 8, 12 }; int N = Arr.length; int K = 2; // Function Call if (findCount(Arr, N, K)) System.out.print("Yes" +"\n"); else System.out.print("No" +"\n");}}// This code contributed by shikhasingrajput |
Python3
# Python code to implement the above approach# Function to check if it is possible# to make even and odd element count same# using at most K operationsdef findCount(arr, N, K): # Edge Case if ((N & 1) != 0): return 0 # Initialize the variables Res = 0 CE = 0 CO = 0 v = [0]*32 # Find the index of rightmost set bit # for every array element and increment # that index of vector, also find even # and odd count for i in range(N): if ((arr[i] & 1) != 0): CO += 1 else: CE += 1 v[arr[i] & (arr[i]-1) - 1] += 1 # Condition is already true if (CE == CO): Res = 0 # Count of Even > Count of Odd elif (CE > CO): n = (CE - CO) / 2 Res = 0 # Find minimum operations to make the # even and odd count equal for i in range(1, len(v)): if (n >= v[i]): Res += (v[i] * i) n = n - v[i] else: Res += n * i break # Count of Odd > Count of Even else: Res = (CO - CE) / 2 return Res <= K# Driver Codeif __name__ == "__main__": Arr = [1, 4, 8, 12] N = len(Arr) K = 2 # Function Call if findCount(Arr, N, K): print("Yes") else: print("No")# This code is contributed by Rohit Pradhan |
C#
// C# code to implement the above approachusing System;public class GFG { // Function to check if it is possible // to make even and odd element count same // using at most K operations static bool findCount(int[] arr, int N, int K) { // Edge Case if ((N & 1) > 0) { return false; } // Initialize the variables int Res, CE = 0, CO = 0; int[] v = new int[32]; // Find the index of rightmost set bit // for every array element and increment // that index of vector, also find even // and odd count for (int i = 0; i < N; i++) { if ((arr[i] & 1) > 0) CO++; else CE++; v[arr[i] & (arr[i] - 1) - 1]++; } // Condition is already true if (CE == CO) { Res = 0; } // Count of Even > Count of Odd else if (CE > CO) { int n = (CE - CO) / 2; Res = 0; // Find minimum operations to make the // even and odd count equal for (int i = 1; i < v.Length; i++) { if (n >= v[i]) { Res += v[i] * i; n = n - v[i]; } else { Res += n * i; break; } } } // Count of Odd > Count of Even else { Res = (CO - CE) / 2; } return Res <= K; } static public void Main() { // Code int[] Arr = { 1, 4, 8, 12 }; int N = Arr.Length; int K = 2; // Function Call if (findCount(Arr, N, K)) Console.Write("Yes" + "\n"); else Console.Write("No" + "\n"); }}// This code contributed by lokeshmvs21. |
Javascript
<script> // JavaScript code to implement the above approach // Function to find rightmost // set bit in given number. function ffs(n) { return Math.log2(n & -n) + 1; } // Function to check if it is possible // to make even and odd element count same // using at most K operations const findCount = (arr, N, K) => { // Edge Case if (N & 1) { return 0; } // Initialize the variables let Res, CE = 0, CO = 0; let v = new Array(32).fill(0); // Find the index of rightmost set bit // for every array element and increment // that index of vector, also find even // and odd count for (let i = 0; i < N; i++) { if (arr[i] & 1) CO++; else CE++; v[ffs(arr[i]) - 1]++; } // Condition is already true if (CE == CO) { Res = 0; } // Count of Even > Count of Odd else if (CE > CO) { let n = parseInt((CE - CO) / 2); Res = 0; // Find minimum operations to make the // even and odd count equal for (let i = 1; i < v.length; i++) { if (n >= v[i]) { Res += v[i] * i; n = n - v[i]; } else { Res += n * i; break; } } } // Count of Odd > Count of Even else { Res = parseInt((CO - CE) / 2); } return Res <= K; } // Driver Code let Arr = [1, 4, 8, 12]; let N = Arr.length; let K = 2; // Function Call if (findCount(Arr, N, K)) document.write("Yes"); else document.write("No"); // This code is contributed by rakeshsahni</script> |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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