Find if it is possible to make a binary string which contains given number of “0”, “1” , “01” and “10” as sub sequences

Given four integers l, m, x, and y. The task is to check whether it is possible to make a binary string consisting of l 0’s, m 1’s, x “01” and y “10” as sub-sequences in it.
Examples:

Input: l = 3, m = 2, x = 4, y = 2 
Output: Yes 
Possible string is “00110”. It contains 3 0’s, 2 1’s, 
4 “01” sub-sequences and 2 “10” sub-sequences.
Input: l = 3, m = 2, x = 4, y = 3 
Output: No 
No such binary string exists. 

Approach: The possible string is always of form 00…11…00…. First consists of some number of zeroes, then all ones, and then the remaining number of zeros. 
Let l1 be the number of zeros before ones and l2 be the number of zeros after ones then the equations are: 

1. l1 + l2 = l (Total number of zeros).
2. l1 * m = x (Number of “01” sub-sequences).
3. m * l2 = y (Number of “10” sub-sequences).

From the above three equations, we get x + y = l * m. If this equation fails for the given values then the string is not possible else print Yes.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if it is possible to
// make a binary string consisting of l 0's, m 1's,
// x "01" sub-sequences and y "10" sub-sequences
bool isPossible(int l, int m, int x, int y)
{
    if (l * m == x + y)
        return true;

    return false;
}

// Driver code
int main()
{
    int l = 3, m = 2, x = 4, y = 2;

    if (isPossible(l, m, x, y))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
class sol
{

// Function that returns true if it is possible to
// make a binary string consisting of l 0's, m 1's,
// x "01" sub-sequences and y "10" sub-sequences
static boolean isPossible(int l, int m, int x, int y)
{
    if (l * m == x + y)
        return true;

    return false;
}

// Driver code
public static void main(String args[])
{
    int l = 3, m = 2, x = 4, y = 2;

    if (isPossible(l, m, x, y))
        System.out.print("Yes");
    else
        System.out.print("No");

}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach

# Function that returns true if it is possible to
# make a binary string consisting of l 0's, m 1's,
# x "01" sub-sequences and y "10" sub-sequences
def isPossible(l, m, x, y) :

    if (l * m == x + y) :
        return True;

    return False;

# Driver code
if __name__ == "__main__" :

    l = 3; m = 2; x = 4; y = 2;

    if (isPossible(l, m, x, y)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class sol
{

// Function that returns true if it is possible to
// make a binary string consisting of l 0's, m 1's,
// x "01" sub-sequences and y "10" sub-sequences
static Boolean isPossible(int l, int m, int x, int y)
{
    if (l * m == x + y)
        return true;

    return false;
}

// Driver code
public static void Main(String []args)
{
    int l = 3, m = 2, x = 4, y = 2;

    if (isPossible(l, m, x, y))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by Arnab Kundu

Javascript

<script>

// Javascript implementation of the approach

// Function that returns true if it is possible to
// make a binary string consisting of l 0's, m 1's,
// x "01" sub-sequences and y "10" sub-sequences
function isPossible(l, m, x, y)
{
    if (l * m == x + y)
        return true;

    return false;
}

// Driver code
    let l = 3, m = 2, x = 4, y = 2;

    if (isPossible(l, m, x, y))
        document.write("Yes");
    else
        document.write("No");

</script>

Output:

Yes

Time Complexity : O(1)

Auxiliary Space: O(1)