Check if all the set bits of the binary representation of N are at least K places away
Given numbers N and K, The task to check if all the set bits of the binary representation of N are at least K places away.
Examples:
Input: N = 5, K = 1 Output: YES Explanation: Binary of 5 is 101. The 1's are 1 place far from each other. Input: N = 10, K = 2 Output: NO Explanation: Binary of 10 is 1010. The 1's are not at least 2 places far from each other.
Approach:
- Iterate Over all the bits in the binary representation of N and maintain a variable ‘count’ initialize to 0.
- Whenever you find a set bit(1), check if count <= K. If not return false.
- If you find a unset bit(0), increases the value of count by 1.
Below is the implementation of the above approach.
// C++ program to check if all the set // bits of the binary representation // of N are at least K places away. #include <bits/stdc++.h> using namespace std; bool CheckBits(int N, int K) { // Initialize check and count // with 0 int check = 0; int count = 0; for (int i = 31; i >= 0; i--) { // The i-th bit is a set bit if ((1 << i) & N) { // This is the first set bit so, // start calculating all the // distances between consecutive // bits from here if (check == 0) { check = 1; } else { // If count is less than K // return false if (count < K) { return false; } } count = 0; } else { // Adding the count as the // number of zeroes increase // between set bits count++; } } return true; } // Driver code int main() { int N = 5; int K = 1; if(CheckBits(N, K)) { cout << "YES"; } else { cout << "NO"; } return 0; } |
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Output:
YES
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