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->4->5.
So the minimum jumps require are 2.Input: N = 23
Output: 4
Explanation:
We can take jumps from 0->1->3->7->23.
So the minimum jumps required are 4.
First 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> |
Output:
4
Time Complexity: O(log N)
Auxiliary Space: O(1)
Second Approach: Since the jumps are required to be in perfect power of 2. We can observe that log2 function gives the highest perfect power of 2 which can be achieved less than N if we typecast it to an integer. So we can subtract the pow(2,(int)log2(N)) each time from N till its value is greater than 0 while incrementing cnt at the same time.
C++14
#include <bits/stdc++.h>using namespace std;int stepRequired(int& N){ int cnt = 0; //until N is reached while(N>0) { //subtract highest perfect power of 2 we can reach from previous level N-=pow(2,(int)log2(N)); //increment cnt for total number of steps taken cnt++; } return cnt;}int main(){ // Number of stairs int N = 23; // Function Call cout << stepRequired(N); return 0;} |
Java
// Java program for above approachimport java.util.*;class GFG{static int stepRequired(int N){ int cnt = 0; //until N is reached while(N>0) { //subtract highest perfect power of 2 we can reach from previous level N-= Math.pow(2, (int)(Math.log(N) / Math.log(2))); //increment cnt for total number of steps taken cnt++; } return cnt;}public static void main(String[] args) { // Number of stairs int N = 23; // Function Call System.out.println(stepRequired(N));}}// This code is contributed by sanjoy_62. |
Python3
# Python code is contributed by shinjanpatraimport mathdef stepRequired(N): cnt = 0 # until N is reached while(N > 0): # subtract highest perfect power of 2 we can reach from previous level N -= math.pow(2,math.floor(math.log2(N))) # increment cnt for total number of steps taken cnt += 1 return cnt# driver code # Number of stairsN = 23 # Function Callprint(stepRequired(N))# This code is contributed by shinjanpatra |
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; //until N is reached while (N > 0) { //subtract highest perfect power of 2 we can reach from previous level N-=(int)Math.Pow(2,(int)(Math.Log(N,2))); //increment cnt for total number of steps taken 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 aditya942003patil |
Javascript
<script>// JavaScript code is contributed by shinjanpatrafunction stepRequired(N){ let cnt = 0; // until N is reached while(N > 0) { // subtract highest perfect power of 2 we can reach from previous level N -= Math.pow(2,Math.floor(Math.log2(N))); // increment cnt for total number of steps taken cnt++; } return cnt;}// driver code // Number of stairslet N = 23; // Function Calldocument.write(stepRequired(N));// This code is contributed by shinjanpatra</script> |
Time Complexity: O(log N)
Auxiliary Space: O(1)
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