Given an array arr[] of size N, the task is to empty given array by removing 2i – 1 array elements in each operation (i is any positive integer). Find the minimum number of operations required.
Examples:
Input: arr[] = { 2, 3, 4 }
Output: 1
Explanation:
Removing (22 – 1) elements i.e { arr[0], arr[1], arr[2] } modifies arr[] to { }
Since no elements left in the array therefore, the required output is 1.Input: arr[] = { 1, 2, 3, 4 }
Output: 2
Explanation:
Removing (21 – 1) element i.e, { arr[0] } modifies arr[] to { 2, 3, 4 }
Removing (22 – 1) elements i.e, { arr[0], arr[1], arr[2] } modifies arr[] to { }
Since no elements left in the array therefore, the required output is 2.
Approach: The problem can be solved using Greedy technique. The idea is to always remove the maximum possible count(2i – 1) of elements from the array. Follow the steps below to solve the problem:
- Initialize a variable, say cntSteps, to store the minimum count of operations required to empty given array.
- Removing N array elements modifies arr[] to 0 length array. Therefore, increment the value of N by 1.
- Traverse each bit of N using variable i and for every ith bit, check if the bit is set or not. If found to be true, then update cntSteps += 1
- Finally, print the value of cntSteps.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find minimum count of steps// required to remove all the array elementsint minimumStepReqArr(int arr[], int N){ // Stores minimum count of steps required // to remove all the array elements int cntStep = 0; // Update N N += 1; // Traverse each bit of N for (int i = 31; i >= 0; i--) { // If current bit is set if (N & (1 << i)) { // Update cntStep cntStep += 1; } } return cntStep;}// Driver Codeint main(){ int arr[] = { 1, 2, 3 }; int N = sizeof(arr) / sizeof(arr[0]); cout << minimumStepReqArr(arr, N); return 0;} |
Java
// Java program to implement // the above approach import java.util.*;class GFG{ // Function to find minimum count of steps // required to remove all the array elements static int minimumStepReqArr(int[] arr, int N) { // Stores minimum count of steps required // to remove all the array elements int cntStep = 0; // Update N N += 1; // Traverse each bit of N for (int i = 31; i >= 0; i--) { // If current bit is set if ((N & (1 << i)) != 0) { // Update cntStep cntStep += 1; } } return cntStep; } // Driver code public static void main(String[] args) { int[] arr = { 1, 2, 3 }; int N = arr.length; System.out.println(minimumStepReqArr(arr, N)); }}// This code is contributed by susmitakundugoaldanga |
Python3
# Python3 program to implement# the above approach# Function to find minimum count of steps# required to remove all the array elementsdef minimumStepReqArr(arr, N): # Stores minimum count of steps required # to remove all the array elements cntStep = 0 # Update N N += 1 i = 31 while(i >= 0): # If current bit is set if (N & (1 << i)): # Update cntStep cntStep += 1 i -= 1 return cntStep# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 3 ] N = len(arr) print(minimumStepReqArr(arr, N))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program to implement // the above approach using System;class GFG { // Function to find minimum count of steps // required to remove all the array elements static int minimumStepReqArr(int[] arr, int N) { // Stores minimum count of steps required // to remove all the array elements int cntStep = 0; // Update N N += 1; // Traverse each bit of N for (int i = 31; i >= 0; i--) { // If current bit is set if ((N & (1 << i)) != 0) { // Update cntStep cntStep += 1; } } return cntStep; } // Driver code static void Main() { int[] arr = { 1, 2, 3 }; int N = arr.Length; Console.WriteLine(minimumStepReqArr(arr, N)); }}// This code is contributed by divyesh072019 |
Javascript
<script> // Javascript program to implement the above approach // Function to find minimum count of steps // required to remove all the array elements function minimumStepReqArr(arr, N) { // Stores minimum count of steps required // to remove all the array elements let cntStep = 0; // Update N N += 1; // Traverse each bit of N for (let i = 31; i >= 0; i--) { // If current bit is set if ((N & (1 << i)) != 0) { // Update cntStep cntStep += 1; } } return cntStep; } let arr = [ 1, 2, 3 ]; let N = arr.length; document.write(minimumStepReqArr(arr, N));// This code is contributed by suresh07.</script> |
Output:
1
Time Complexity: O(31)
Auxiliary Space: O(1)
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