Given two non-negative integers m and k. The problem is to find the m-th smallest number having k number of set bits.
Constraints: 1 <= m, k.
Examples:
Input : m = 4, k = 2 Output : 9 (9)10 = (1001)2, it is the 4th smallest number having 2 set bits. Input : m = 6, k = 4 Output : 39
Approach: Following are the steps:
- Find the smallest number having k number of set bits. Let it be num, where num = (1 << k) – 1.
- Loop for m-1 times and each time replace num with the next higher number than ‘num’ having same number of bits as in ‘num’. Refer this post to find the required next higher number.
- Finally return num.
C++
// C++ implementation to find the mth smallest // number having k number of set bits #include <bits/stdc++.h> using namespace std; typedef unsigned int uint_t; // function to find the next higher number // with same number of set bits as in 'x' uint_t nxtHighWithNumOfSetBits(uint_t x) { uint_t rightOne; uint_t nextHigherOneBit; uint_t rightOnesPattern; uint_t next = 0; /* the approach is same as dicussed in */ if (x) { rightOne = x & -(signed)x; nextHigherOneBit = x + rightOne; rightOnesPattern = x ^ nextHigherOneBit; rightOnesPattern = (rightOnesPattern) / rightOne; rightOnesPattern >>= 2; next = nextHigherOneBit | rightOnesPattern; } return next; } // function to find the mth smallest number // having k number of set bits int mthSmallestWithKSetBits(uint_t m, uint_t k) { // smallest number having 'k' // number of set bits uint_t num = (1 << k) - 1; // finding the mth smallest number // having k set bits for (int i = 1; i < m; i++) num = nxtHighWithNumOfSetBits(num); // required number return num; } // Driver program to test above int main() { uint_t m = 6, k = 4; cout << mthSmallestWithKSetBits(m, k); return 0; } |
Python3
# Python3 implementation to find the mth # smallest number having k number of set bits # function to find the next higher number # with same number of set bits as in 'x' def nxtHighWithNumOfSetBits(x): rightOne = 0 nextHigherOneBit = 0 rightOnesPattern = 0 next = 0 """ the approach is same as dicussed in http:#www.geeksforgeeks.org/next-higher-number-with-same-number-of-set-bits/ """ if (x): rightOne = x & (-x) nextHigherOneBit = x + rightOne rightOnesPattern = x ^ nextHigherOneBit rightOnesPattern = (rightOnesPattern) // rightOne rightOnesPattern >>= 2 next = nextHigherOneBit | rightOnesPattern return next # function to find the mth smallest # number having k number of set bits def mthSmallestWithKSetBits(m, k): # smallest number having 'k' # number of set bits num = (1 << k) - 1 # finding the mth smallest number # having k set bits for i in range(1, m): num = nxtHighWithNumOfSetBits(num) # required number return num # Driver Code if __name__ == '__main__': m = 6 k = 4 print(mthSmallestWithKSetBits(m, k)) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
Output:
39
Time Complexity: O(m)
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