Count the maximum inversion count by concatenating the given Strings

Given A, the number of "1" strings, B number of "10" strings, and C number of "0" strings. The task is to count the maximum inversion count by concatenating these strings

Note: Inversion count is defined as the number of pairs (i, j) such that 0 ≤ i < j ≤ N-1 and S[i] = '1' and S[j] = '0'.

Examples:

Input: A = 2, B = 1, C = 0
Output: 3
Explanation: Optimal string = "1110", hence total number of inversions is 3.

Input: A = 0, B = 0, C = 1
Output: 0
Explanation: Only possible string = "0", hence total number of inversions is 0.

Approach: This can be solved with the following idea:

It is always optimal to include A and B strings in our answer. Try to form maximum strings from A and B, Increase the inversion by concatinating C at last. As it always contains 0 in it's string.

Below are the steps involved:

• Initialize a integer ans = 0.
• Form A * (B + C), add it to ans.
• Then form B and C, by B * C add it to ans.
• Try forming the ones with B to increase inversion.
• Return ans.

Below is the implementation of the code:

// C++ code for the above approach:
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to count maximum Inversion
int maxInversion(int A, int B, int C)
{

    // Forming ABC
    long long ans = A * 1LL * (B + C);

    // Forming BC
    ans += (B * 1LL * C);

    // Checking all Pairs possible from B
    ans += (B * 1LL * (B + 1) / 2);

    // Return total count
    return ans;
}

// Driver Code
int main()
{

    int A = 2;
    int B = 1;
    int C = 0;

    // Function call
    cout << maxInversion(A, B, C);
    return 0;
}

// Java Implementation

public class Main {
    public static void main(String[] args) {
        int A = 2;
        int B = 1;
        int C = 0;

        // Function call
        System.out.println(maxInversion(A, B, C));
    }

    // Function to count maximum Inversion
    public static long maxInversion(int A, int B, int C) {

        // Forming ABC
        long ans = A * (long) (B + C);

        // Forming BC
        ans += (B * (long) C);

        // Checking all Pairs possible from B
        ans += (B * (long) (B + 1) / 2);

        // Return total count
        return ans;
    }
}

// This code is contributed by Sakshi

def max_inversion(A, B, C):
    # Forming ABC
    ans = A * (B + C)

    # Forming BC
    ans += B * C

    # Checking all Pairs possible from B
    ans += B * (B + 1) // 2

    # Return total count
    return ans

# Driver Code
if __name__ == "__main__":
    A = 2
    B = 1
    C = 0

    # Function call
    print(max_inversion(A, B, C))

using System;

class Program
{
    // Function to count maximum Inversion
    static long MaxInversion(int A, int B, int C)
    {
        // Forming ABC
        long ans = A * 1L * (B + C);

        // Forming BC
        ans += B * 1L * C;

        // Checking all Pairs possible from B
        ans += B * 1L * (B + 1) / 2;

        // Return total count
        return ans;
    }

    // Driver Code
    static void Main()
    {
        int A = 2;
        int B = 1;
        int C = 0;

        // Function call
        Console.WriteLine(MaxInversion(A, B, C));
    }
}

function GFG(A, B, C) {
    // Forming ABC
    let ans = A * (B + C);
    // Forming BC
    ans += B * C;
    // Checking all pairs possible from B
    ans += (B * (B + 1)) / 2;
    // Return total count
    return ans;
}
// Driver Code
function main() {
    // Given values
    const A = 2;
    const B = 1;
    const C = 0;
    // Function call
    console.log(GFG(A, B, C));
}
main();

3

Time Complexity: O(1)
Auxiliary Space: O(1)

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Article Tags :

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• Geeks Premier League
• DSA
• Data Structures
• Geeks Premier League 2023

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