Given two numbers x, y which denotes the number of set bits. Also given is a number C. The task is to print the number of ways in which we can form two numbers A and B such that A has x number of set bits and B has y number of set bits and A+B = C.
Examples:
Input: X = 1, Y = 1, C = 3 Output: 2 So two possible ways are (A = 2 and B = 1) and (A = 1 and B = 2) Input: X = 2, Y = 2, C = 20 Output: 3
Recursive Approach: The above problem can be solved using recursion. Calculate the total count of pairs recursively. There will have 4 possibilities. We start from the rightmost bit position.
- If the bit position at C is 1, then there are four possibilities to get 1 at that index.
- If the carry is 0, then we can use 1 bit out of X and 0 bit out of Y, or the vice-versa which generates no carry for next step.
- If the carry is 1, then we can use 1 set bit’s from each which generates a carry for the next step else we use no set bits from X and Y which generates no carry.
- If the bit position at C is 0, then there are four possibilities to get 0 at that bit position.
- If the carry is 1, then we can use 1 set bit out of X and 0 bit out of Y or the vice versa which generates a carry of 1 for the next step.
- If the carry is 0, then we can use 1 and 1 out of X and Y respectively, which generates a carry of 1 for next step. We can also use no set bits, which generates no carry for the next step.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Recursive function to generate// all combinations of bitsint func(int third, int seta, int setb, int carry, int number){ // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 and seta == 0 and setb == 0 and carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 or seta < 0 or setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = shift & 1; int ans = 0; // carry = 1 and bit set = 1 if ((mask) && carry) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask && !carry) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (!mask && carry) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (!mask && !carry) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // Return ans return ans;}// Function to count waysint possibleSwaps(int a, int b, int c){ // Call func return func(0, a, b, 0, c);}// Driver Codeint main(){ int x = 2, y = 2, c = 20; cout << possibleSwaps(x, y, c); return 0;} |
Java
// Java implementation of above approachimport java.util.*;public class GFG{// Recursive function to generate// all combinations of bitsstatic int func(int third, int seta, int setb, int carry, int number){ // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = (shift & 1); int ans = 0; // carry = 1 and bit set = 1 if ((mask) == 1 && carry == 1) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // Return ans return ans;}// Function to count waysstatic int possibleSwaps(int a, int b, int c){ // Call func return func(0, a, b, 0, c);}// Driver Codepublic static void main(String args[]){ int x = 2, y = 2, c = 20; System.out.println(possibleSwaps(x, y, c));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python implementation of above approach# Recursive function to generate# all combinations of bitsdef func(third, seta, setb, carry, number): # find if C has no more set bits on left shift = (number >> third); # if no set bits are left for C # and there are no set bits for A and B # and the carry is 0, then # this combination is possible if (shift == 0 and seta == 0 and setb == 0 and carry == 0): return 1; # if no set bits are left for C and # requirement of set bits for A and B have exceeded if (shift == 0 or seta < 0 or setb < 0): return 0; # Find if the bit is 1 or 0 at # third index to the left mask = (shift & 1); ans = 0; # carry = 1 and bit set = 1 if ((mask) == 1 and carry == 1): # since carry is 1, and we need 1 at C's bit position # we can use 0 and 0 # or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); # carry = 0 and bit set = 1 elif(mask == 1 and carry == 0): # since carry is 0, and we need 1 at C's bit position # we can use 1 and 0 # or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); # carry = 1 and bit set = 0 elif(mask == 0 and carry == 1): # since carry is 1, and we need 0 at C's bit position # we can use 1 and 0 # or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); # carry = 0 and bit set = 0 elif(mask == 0 and carry == 0): # since carry is 0, and we need 0 at C's bit position # we can use 0 and 0 # or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); # Return ans return ans;# Function to count waysdef possibleSwaps(a, b, c): # Call func return func(0, a, b, 0, c);# Driver Codeif __name__ == '__main__': x = 2; y = 2; c = 20; print(possibleSwaps(x, y, c));# This code is contributed by Rajput-Ji |
C#
// C# implementation of above approachusing System;public class GFG { // Recursive function to generate // all combinations of bits static int func(int third, int seta, int setb, int carry, int number) { // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = (shift & 1); int ans = 0; // carry = 1 and bit set = 1 if ((mask) == 1 && carry == 1) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // Return ans return ans; } // Function to count ways static int possibleSwaps(int a, int b, int c) { // Call func return func(0, a, b, 0, c); } // Driver Code public static void Main(String []args) { int x = 2, y = 2, c = 20; Console.WriteLine(possibleSwaps(x, y, c)); }}// This code is contributed by umadevi9616 |
Javascript
<script>// javascript implementation of above approach // Recursive function to generate // all combinations of bits function func(third , seta , setb , carry , number) { // find if C has no more set bits on left var shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left var mask = (shift & 1); var ans = 0; // carry = 1 and bit set = 1 if ((mask) == 1 && carry == 1) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // Return ans return ans; } // Function to count ways function possibleSwaps(a , b , c) { // Call func return func(0, a, b, 0, c); } // Driver Code var x = 2, y = 2, c = 20; document.write(possibleSwaps(x, y, c));// This code is contributed by Rajput-Ji</script> |
Output
3
Efficient Approach: The above problem can be solved using bitmask DP.
- Initialize a 4-D DP array of size 64 * 64 * 64 * 2 as 10^18 has a maximum of 64 set bits with -1
- The first state of the DP array stores the number of bits traversed in C from right. The second state stores the number of set-bits used out of X and third state stores the number of set bits used out of Y. The fourth state is the carry bit which refers to the carry generated when we perform an addition operation.
- The recurrence will have 4 possibilities. We start from the rightmost bit position.
- If the bit position at C is 1, then there are four possibilities to get 1 at that index.
- If the carry is 0, then we can use 1 bit out of X and 0 bit out of Y, or the vice-versa which generates no carry for next step.
- If the carry is 1, then we can use 1 set bit’s from each which generates a carry for the next step else we use no set bits from X and Y which generates no carry.
- If the bit position at C is 0, then there are four possibilities to get 0 at that bit position.
- If the carry is 1, then we can use 1 set bit out of X and 0 bit out of Y or the vice versa which generates a carry of 1 for the next step.
- If the carry is 0, then we can use 1 and 1 out of X and Y respectively, which generates a carry of 1 for next step. We can also use no set bits, which generates no carry for the next step.
- Summation of all possibilities is stored in dp[third][seta][setb][carry] to avoid re-visiting same states.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Initial DP arrayint dp[64][64][64][2];// Recursive function to generate// all combinations of bitsint func(int third, int seta, int setb, int carry, int number){ // if the state has already been visited if (dp[third][seta][setb][carry] != -1) return dp[third][seta][setb][carry]; // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 and seta == 0 and setb == 0 and carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 or seta < 0 or setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = shift & 1; dp[third][seta][setb][carry] = 0; // carry = 1 and bit set = 1 if ((mask) && carry) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask && !carry) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (!mask && carry) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (!mask && !carry) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } return dp[third][seta][setb][carry];}// Function to count waysint possibleSwaps(int a, int b, int c){ memset(dp, -1, sizeof(dp)); // function call that returns the // answer int ans = func(0, a, b, 0, c); return ans;}// Driver Codeint main(){ int x = 2, y = 2, c = 20; cout << possibleSwaps(x, y, c); return 0;} |
Java
// Java implementation of the above approach import java.util.*;class GfG { // Initial DP array static int dp[][][][] = new int[64][64][64][2]; // Recursive function to generate // all combinations of bits static int func(int third, int seta, int setb, int carry, int number) { // if the state has already been visited if (dp[third][seta][setb][carry] != -1) return dp[third][seta][setb][carry]; // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = shift & 1; dp[third][seta][setb][carry] = 0; // carry = 1 and bit set = 1 if ((mask == 1) && carry == 1) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third][seta][setb][carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } return dp[third][seta][setb][carry]; } // Function to count ways static int possibleSwaps(int a, int b, int c) { for(int q = 0; q < 64; q++) { for(int r = 0; r < 64; r++) { for(int p = 0; p < 64; p++) { for(int d = 0; d < 2; d++) { dp[q][r][p][d] = -1; } } } } // function call that returns the // answer int ans = func(0, a, b, 0, c); return ans; } // Driver Code public static void main(String[] args) { int x = 2, y = 2, c = 20; System.out.println(possibleSwaps(x, y, c)); }} |
Python3
# Python3 implementation of the above approach # Initial DP array dp = [[[[-1, -1] for i in range(64)] for j in range(64)] for k in range(64)] # Recursive function to generate # all combinations of bits def func(third, seta, setb, carry, number): # if the state has already been visited if dp[third][seta][setb][carry] != -1: return dp[third][seta][setb][carry] # find if C has no more set bits on left shift = number >> third # if no set bits are left for C # and there are no set bits for A and B # and the carry is 0, then # this combination is possible if (shift == 0 and seta == 0 and setb == 0 and carry == 0): return 1 # if no set bits are left for C and # requirement of set bits for A and B have exceeded if (shift == 0 or seta < 0 or setb < 0): return 0 # Find if the bit is 1 or 0 at # third index to the left mask = shift & 1 dp[third][seta][setb][carry] = 0 # carry = 1 and bit set = 1 if (mask) and carry: # since carry is 1, and we need 1 at # C's bit position we can use 0 and 0 # or 1 and 1 at A and B bit position dp[third][seta][setb][carry] +=\ func(third + 1, seta, setb, 0, number) + \ func(third + 1, seta - 1, setb - 1, 1, number) # carry = 0 and bit set = 1 elif mask and not carry: # since carry is 0, and we need 1 at C's # bit position we can use 1 and 0 # or 0 and 1 at A and B bit position dp[third][seta][setb][carry] +=\ func(third + 1, seta - 1, setb, 0, number) + \ func(third + 1, seta, setb - 1, 0, number) # carry = 1 and bit set = 0 elif not mask and carry: # since carry is 1, and we need 0 at C's # bit position we can use 1 and 0 # or 0 and 1 at A and B bit position dp[third][seta][setb][carry] +=\ func(third + 1, seta - 1, setb, 1, number) + \ func(third + 1, seta, setb - 1, 1, number) # carry = 0 and bit set = 0 elif not mask and not carry: # since carry is 0, and we need 0 at C's # bit position we can use 0 and 0 # or 1 and 1 at A and B bit position dp[third][seta][setb][carry] += \ func(third + 1, seta, setb, 0, number) + \ func(third + 1, seta - 1, setb - 1, 1, number) return dp[third][seta][setb][carry] # Function to count ways def possibleSwaps(a, b, c): # function call that returns the answer ans = func(0, a, b, 0, c) return ans # Driver Code if __name__ == "__main__": x, y, c = 2, 2, 20 print(possibleSwaps(x, y, c)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the above approach using System;class GFG { // Initial DP array static int [,,,]dp = new int[64, 64, 64, 2]; // Recursive function to generate // all combinations of bits static int func(int third, int seta, int setb, int carry, int number) { // if the state has already been visited if (third > -1 && seta > -1 && setb > -1 && carry > -1) if(dp[third, seta, setb, carry] != -1) return dp[third, seta, setb, carry]; // find if C has no more set bits on left int shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and // B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left int mask = shift & 1; dp[third, seta, setb, carry] = 0; // carry = 1 and bit set = 1 if ((mask == 1) && carry == 1) { // since carry is 1, and we need 1 at // C's bit position, we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third, seta, setb, carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at // C's bit position, we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third, seta, setb, carry] += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at // C's bit position, we can use 1 and 0 // or 0 and 1 at A and B bit position dp[third, seta, setb, carry] += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at // C's bit position, we can use 0 and 0 // or 1 and 1 at A and B bit position dp[third, seta, setb, carry] += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } return dp[third, seta, setb, carry]; } // Function to count ways static int possibleSwaps(int a, int b, int c) { for(int q = 0; q < 64; q++) { for(int r = 0; r < 64; r++) { for(int p = 0; p < 64; p++) { for(int d = 0; d < 2; d++) { dp[q, r, p, d] = -1; } } } } // function call that returns the // answer int ans = func(0, a, b, 0, c); return ans; } // Driver Code public static void Main(String[] args) { int x = 2, y = 2, c = 20; Console.WriteLine(possibleSwaps(x, y, c)); }} // This code is contributed by Rajput-Ji |
Javascript
<script>// javascript implementation of above approach // Recursive function to generate // all combinations of bits function func(third , seta , setb , carry , number) { // find if C has no more set bits on left var shift = (number >> third); // if no set bits are left for C // and there are no set bits for A and B // and the carry is 0, then // this combination is possible if (shift == 0 && seta == 0 && setb == 0 && carry == 0) return 1; // if no set bits are left for C and // requirement of set bits for A and B have exceeded if (shift == 0 || seta < 0 || setb < 0) return 0; // Find if the bit is 1 or 0 at // third index to the left var mask = (shift & 1); var ans = 0; // carry = 1 and bit set = 1 if ((mask) == 1 && carry == 1) { // since carry is 1, and we need 1 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // carry = 0 and bit set = 1 else if (mask == 1 && carry == 0) { // since carry is 0, and we need 1 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 0, number) + func(third + 1, seta, setb - 1, 0, number); } // carry = 1 and bit set = 0 else if (mask == 0 && carry == 1) { // since carry is 1, and we need 0 at C's bit position // we can use 1 and 0 // or 0 and 1 at A and B bit position ans += func(third + 1, seta - 1, setb, 1, number) + func(third + 1, seta, setb - 1, 1, number); } // carry = 0 and bit set = 0 else if (mask == 0 && carry == 0) { // since carry is 0, and we need 0 at C's bit position // we can use 0 and 0 // or 1 and 1 at A and B bit position ans += func(third + 1, seta, setb, 0, number) + func(third + 1, seta - 1, setb - 1, 1, number); } // Return ans return ans; } // Function to count ways function possibleSwaps(a , b , c) { // Call func return func(0, a, b, 0, c); } // Driver Code var x = 2, y = 2, c = 20; document.write(possibleSwaps(x, y, c));// This code contributed by Rajput-Ji </script> |
Output:
3
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