Given a number N, the task is to find the maximum distance between two 1’s in the binary representation of given N. Print -1 if binary representation contains less than two 1’s.
Examples:
Input: N = 131 Output: 7 131 in binary = 10000011. The maximum distance between two 1's = 7. Input: N = 8 Output: -1 8 in binary = 01000. It contains less than two 1's.
Approach:
- First find the binary representation of N.
- For each bit calculated, check if its a ‘1’.
- Store the index of first ‘1’ found in first_1, and the last ‘1’ found in last_1
- Then check if the last_1 is less than or equal to first_1. It will be the case when N is a power of 2. Hence print -1 in this case.
- In any other case, find the difference between the last_1 and first_1. This will be the required distance.
Below is the implementation of the above approach:
C++
// C++ program to find the// Maximum distance between two 1's// in Binary representation of N#include <bits/stdc++.h>using namespace std;int longest_gap(int N){ int distance = 0, count = 0, first_1 = -1, last_1 = -1; // Compute the binary representation while (N) { count++; int r = N & 1; if (r == 1) { first_1 = first_1 == -1 ? count : first_1; last_1 = count; } N = N / 2; } // if N is a power of 2 // then return -1 if (last_1 <= first_1) { return -1; } // else find the distance // between the first position of 1 // and last position of 1 else { distance = (last_1 - first_1); return distance; }}// Driver codeint main(){ int N = 131; cout << longest_gap(N) << endl; N = 8; cout << longest_gap(N) << endl; N = 17; cout << longest_gap(N) << endl; N = 33; cout << longest_gap(N) << endl; return 0;} |
Java
// Java program to find the // Maximum distance between two 1's // in Binary representation of N class GFG{ static int longest_gap(int N) { int distance = 0, count = 0, first_1 = -1, last_1 = -1; // Compute the binary representation while (N != 0) { count++; int r = N & 1; if (r == 1) { first_1 = first_1 == -1 ? count : first_1; last_1 = count; } N = N / 2; } // if N is a power of 2 // then return -1 if (last_1 <= first_1) { return -1; } // else find the distance // between the first position of 1 // and last position of 1 else { distance = (last_1 - first_1); return distance; } } // Driver code public static void main (String[] args) { int N = 131; System.out.println(longest_gap(N)); N = 8; System.out.println(longest_gap(N)); N = 17; System.out.println(longest_gap(N)); N = 33; System.out.println(longest_gap(N)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 program to find the# Maximum distance between two 1's# in Binary representation of Ndef longest_gap(N): distance = 0 count = 0 first_1 = -1 last_1 = -1 # Compute the binary representation while (N > 0): count += 1 r = N & 1 if (r == 1): if first_1 == -1: first_1 = count else: first_1 = first_1 last_1 = count N = N // 2 # if N is a power of 2 # then return -1 if (last_1 <= first_1): return -1 # else find the distance # between the first position of 1 # and last position of 1 else: distance = last_1 - first_1 return distance# Driver codeN = 131print(longest_gap(N))N = 8print(longest_gap(N))N = 17print(longest_gap(N))N = 33print(longest_gap(N))# This code is contributed by Mohit Kumar |
C#
// C# program to find the // Maximum distance between two 1's // in Binary representation of N using System;class GFG{ static int longest_gap(int N) { int distance = 0, count = 0, first_1 = -1, last_1 = -1; // Compute the binary representation while (N != 0) { count++; int r = N & 1; if (r == 1) { first_1 = first_1 == -1 ? count : first_1; last_1 = count; } N = N / 2; } // if N is a power of 2 // then return -1 if (last_1 <= first_1) { return -1; } // else find the distance // between the first position of 1 // and last position of 1 else { distance = (last_1 - first_1); return distance; } } // Driver code public static void Main (String []args) { int N = 131; Console.WriteLine(longest_gap(N)); N = 8; Console.WriteLine(longest_gap(N)); N = 17; Console.WriteLine(longest_gap(N)); N = 33; Console.WriteLine(longest_gap(N)); } }// This code is contributed by Arnab Kundu |
Javascript
<script>// Javascript program to find the// Maximum distance between two 1's// in Binary representation of Nfunction longest_gap(N){ let distance = 0, count = 0, first_1 = -1, last_1 = -1; // Compute the binary representation while (N) { count++; let r = N & 1; if (r == 1) { first_1 = first_1 == -1 ? count : first_1; last_1 = count; } N = parseInt(N / 2); } // if N is a power of 2 // then return -1 if (last_1 <= first_1) { return -1; } // else find the distance // between the first position of 1 // and last position of 1 else { distance = (last_1 - first_1); return distance; }}// Driver code let N = 131; document.write(longest_gap(N) + "<br>"); N = 8; document.write(longest_gap(N) + "<br>"); N = 17; document.write(longest_gap(N) + "<br>"); N = 33; document.write(longest_gap(N) + "<br>");</script> |
Output
7 -1 4 5
Time Complexity: O(log N)
Auxiliary Space: O(1)
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