Minimum steps to reach the Nth stair in jumps of perfect power of 2
Given N stairs, the task is to find the minimum number of jumps of perfect power of 2 requires to reach the Nth stair.
Examples:
Input: N = 5
Output:
Explanation:
We can take jumps from 0->1->4.
So the minimum jumps require are 2.Input: N = 23
Output: 4
Explanation:
We can take jumps from 0->1->2->4->16.
So the minimum jumps required are 4.
Approach: Since the jumps are required to be in perfect power of 2. So the count of set bit in the given number N is the minimum number of jumps required to reach Nth stair as the summation of 2i for all set bit index i is equals to N.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include "bits/stdc++.h"using namespace std;// Function to count the number of jumps// required to reach Nth stairs.int stepRequired(int N){ int cnt = 0; // Till N becomes 0 while (N) { // Removes the set bits from // the right to left N = N & (N - 1); cnt++; } return cnt;}// Driver Codeint main(){ // Number of stairs int N = 23; // Function Call cout << stepRequired(N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to count the number of jumps// required to reach Nth stairs.static int stepRequired(int N){ int cnt = 0; // Till N becomes 0 while (N > 0) { // Removes the set bits from // the right to left N = N & (N - 1); cnt++; } return cnt;}// Driver Codepublic static void main(String[] args){ // Number of stairs int N = 23; // Function Call System.out.print(stepRequired(N));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program for the above approach# Function to count the number of jumps# required to reach Nth stairs.def stepRequired(N): cnt = 0; # Till N becomes 0 while (N > 0): # Removes the set bits from # the right to left N = N & (N - 1); cnt += 1; return cnt;# Driver Codeif __name__ == '__main__': # Number of stairs N = 23; # Function Call print(stepRequired(N)); # This code is contributed by 29AjayKumar |
C#
// C# program for the above approachusing System;class GFG{// Function to count the number of// jumps required to reach Nth stairs.static int stepRequired(int N){ int cnt = 0; // Till N becomes 0 while (N > 0) { // Removes the set bits from // the right to left N = N & (N - 1); cnt++; } return cnt;}// Driver Codepublic static void Main(String[] args){ // Number of stairs int N = 23; // Function Call Console.Write(stepRequired(N));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the above approach// Function to count the number of jumps// required to reach Nth stairs.function stepRequired(N){ let cnt = 0; // Till N becomes 0 while (N) { // Removes the set bits from // the right to left N = N & (N - 1); cnt++; } return cnt;}// Driver Code// Number of stairslet N = 23;// Function Calldocument.write(stepRequired(N));// This code is contributed by Surbhi Tyagi</script> |
4
Time Complexity: O(log N)
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