# Count pairs (A, B) such that A has X and B has Y number of set bits and A+B = C

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 = 3Output:2 So two possible ways are (A = 2 and B = 1) and (A = 1 and B = 2)Input:X = 2, Y = 2, C = 20Output:3

**Approach:** The above problem can be solved using bitmask DP.

- Initialise 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 possibilites 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 array ` `int` `dp[64][64][64][2]; ` ` ` `// Recursive function to generate ` `// all combinations of bits ` `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 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 ways ` `int` `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 Code ` `int` `main() ` `{ ` ` ` ` ` `int` `x = 2, y = 2, c = 20; ` ` ` ` ` `cout << possibleSwaps(x, y, c); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

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## 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)); ` `} ` `} ` |

*chevron_right*

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## 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
[tabbyending]

**Output:**

3

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