Given two positive integers A and B, we can change at most K bits in both the numbers to make OR of them equal to a given target number T. In the case of multiple solutions try to keep A as small as possible.

Examples:

Input : A = 175, B = 66, T = 100, K = 5 Output : A = 36 B = 64 Initial bits of A = 1010 1111 Changed bits of A = 0010 0100 Initial bits of B = 0100 0010 Changed bits of B = 0100 0000 OR of changed Bits = 0110 0100 Which has decimal value equal to Target T. Input : A = 175, B = 66, T = 100, K = 4 Output : Not Possible It is not possible to get OR of A and B as T, just by changing K bits.

We can solve this problem by iterating over all bits of A and B and greedily changing them that is,

- If i-th bit of Target T is 0 then set i-th bits of A and B to 0 (if not already)
- If i-th bit of Target T is 1 then we will try to set one of the bits to 1 and we will change i-th bit of B only to 1(if not already) to minimize A.

After above procedure, if **changed bits are more than K**, then it is not possible to get OR of A and B as T by changing at most K bits.

If **changed bits are less than k**, then we can further minimize the value of A by using remaining value of K for which we will loop over bits one more time and if at any time,

- i-th A bit is 1 and i-th B bit is 0 then we will make 2 changes and flip both.
- i-th A and B bits are 1 then again we will make 1 change and flip A’s bit.

Total time complexity of above solution will be O(max number of bits).

// C++ program to change least bits to // get desired OR value #include <bits/stdc++.h> using namespace std; // Returns max of three numbers int max(int a, int b, int c) { return max(a, max(b, c)); } // Returns count of bits in N int bitCount(int N) { int cnt = 0; while (N) { cnt++; N >>= 1; } return cnt; } // Returns bit at 'pos' position bool at_position(int num, int pos) { bool bit = num & (1<<pos); return bit; } // Utility method to toggle bit at // 'pos' position void toggle(int &num,int pos) { num ^= (1 << pos); } // method returns minimum number of bit flip // to get T as OR value of A and B void minChangeToReachTaregetOR(int A, int B, int K, int T) { int maxlen = max(bitCount(A), bitCount(B), bitCount(T)); // Loop over maximum number of bits among // A, B and T for (int i = maxlen - 1; i >= 0; i--) { bool bitA = at_position(A, i); bool bitB = at_position(B, i); bool bitT = at_position(T, i); // T's bit is set, try to toggle bit // of B, if not already if (bitT) { if (!bitA && !bitB) { toggle(B, i); K--; } } else { // if A's bit is set, flip that if (bitA) { toggle(A, i); K--; } // if B's bit is set, flip that if (bitB) { toggle(B, i); K--; } } } // if K is less than 0 then we can make A|B == T if (K < 0) { cout << "Not possible\n"; return; } // Loop over bits one more time to minimise // A further for (int i = maxlen - 1; K > 0 && i >= 0; --i) { bool bitA = at_position(A, i); bool bitB = at_position(B, i); bool bitT = at_position(T, i); if (bitT) { // If both bit are set, then Unset // A's bit to minimise it if (bitA && bitB) { toggle(A, i); K--; } } // If A's bit is 1 and B's bit is 0, // toggle both if (bitA && !bitB && K >= 2) { toggle(A, i); toggle(B, i); K -= 2; } } // Output changed value of A and B cout << A << " " << B << endl; } // Driver code int main() { int A = 175, B = 66, K = 5, T = 100; minChangeToReachTaregetOR(A, B, K, T); return 0; }

Output:

36 64

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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