Identify the Most Strongest Programmer

There are N Players in a Game and you have given M pieces of information about their superiority. The i^th information ( infromation[i] = {A_i, B_i} ) states that "Player A_i > Player B_i", the task is to determine the number of the strongest programmers among the N players based on the provided information. If there is a unique strongest programmer, print their number. Otherwise, if there are multiple possible strongest programmers or cannot be determined, print -1.

Examples:

Input: N = 3, M = 2, information[] = [[1, 2], [2, 3]]
Output: 1

Input: N = 6, M = 6, information[] = [[1, 6], [6, 5], [6, 2], [2, 3], [4, 3], [4, 2]]
Output: -1

Approach: To solve the problem follow the below idea:

This problem can be solved using transitive property and some observations.

• If there exists a player for which there are no other players stronger than him then he will be the strongest player.
• To efficiently determine the strongest player among N competitors, We count how many people are stronger than each player.
• It iterates through the given information, updating the count based on the transitive nature of superiority.
• It then checks if there is exactly one candidate with zero stronger players, indicating the strongest person.
• If there are multiple candidates, it cannot identify the Most strongest player in that case we return -1.

Steps to solve the problem:

• Read the input values N (the number of programmers) and M (the number of pieces of information).
• Create a vector deg[] to keep track of the count of people stronger than each person i. Initialize all elements of deg[] to 0.
• For each pair of information (A[i], B[i]) in the given M pieces of information, This step is equivalent to adding 1 to the count of people stronger than each programmer A[i].
• Initialize the answer variable ans to -1, which will store the result.
• Iterate over all the player from 1 to N.
• If there is a player i having deg[i] = 0, then assign i to strongest player and if strongest programmer is not -1 , i.e. there exist another stronger programmer then print -1 and exit the code.
• In the end print strongest programmer

Below is the implementation of above approach:

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

// Function to find the strongest programmer
int findStrongestProgrammer(int N, int M,
                        vector<pair<int, int> > power)
{
    vector<int> deg(N + 1, 0);

    // Taking indegree
    for (int i = 0; i < M; i++) {
        int a = power[i].first;
        int b = power[i].second;
        deg[b]++;
    }
    int strongest_programmer = -1;

    // Finding the strongest programmer
    for (int i = 1; i <= N; i++) {
        if (deg[i] == 0) {
            if (strongest_programmer != -1) {
                return -1;
            }
            else {
                strongest_programmer = i;
            }
        }
    }
    return strongest_programmer;
}

// Drivers code
int main()
{
    int N = 3, M = 2;
    vector<pair<int, int> > power = { { 1, 2 }, { 2, 3 } };

    // Function Call
    int result = findStrongestProgrammer(N, M, power);

    if (result == -1) {
        cout << -1 << endl;
    }
    else {
        cout << result << endl;
    }
    return 0;
}

public class GFG {
    
    static int findStrongestProgrammer(int N, int M, int[][] power) {
        int[] deg = new int[N + 1];

        // Taking indegree
        for (int i = 0; i < M; i++) {
            int b = power[i][1];
            deg[b]++;
        }

        int strongestProgrammer = -1;

        // Finding the strongest programmer
        for (int i = 1; i <= N; i++) {
            if (deg[i] == 0) {
                if (strongestProgrammer != -1) {
                    return -1;
                } else {
                    strongestProgrammer = i;
                }
            }
        }

        return strongestProgrammer;
    }

    public static void main(String[] args) {
        int N = 3;
        int M = 2;
        int[][] power = new int[M][];
        power[0] = new int[]{1, 2};
        power[1] = new int[]{2, 3};

        // Function Call
        int result = findStrongestProgrammer(N, M, power);

        if (result == -1) {
            System.out.println(-1);
        } else {
            System.out.println(result);
        }
    }
}
//this code is contributed by rohit singh

# Function to find the strongest programmer
def findStrongestProgrammer(N, M, power):
    deg = [0] * (N + 1)

    # Taking indegree
    for i in range(M):
        a, b = power[i]
        deg[b] += 1

    strongest_programmer = -1

    # Finding the strongest programmer
    for i in range(1, N + 1):
        if deg[i] == 0:
            if strongest_programmer != -1:
                return -1
            else:
                strongest_programmer = i

    return strongest_programmer

# Driver code
if __name__ == "__main__":
    N = 3
    M = 2
    power = [(1, 2), (2, 3)]

    # Function Call
    result = findStrongestProgrammer(N, M, power)

    if result == -1:
        print(-1)
    else:
        print(result)

#Contributed by Aditi Tyagi

//C# code for the above approach:

using System;

class Program
{
    static int FindStrongestProgrammer(int N, int M, int[][] power)
    {
        int[] deg = new int[N + 1];

        // Taking indegree
        for (int i = 0; i < M; i++)
        {
            int b = power[i][1];
            deg[b]++;
        }

        int strongestProgrammer = -1;

        // Finding the strongest programmer
        for (int i = 1; i <= N; i++)
        {
            if (deg[i] == 0)
            {
                if (strongestProgrammer != -1)
                {
                    return -1;
                }
                else
                {
                    strongestProgrammer = i;
                }
            }
        }

        return strongestProgrammer;
    }

    static void Main()
    {
        int N = 3;
        int M = 2;
        int[][] power = new int[M][];
        power[0] = new int[] { 1, 2 };
        power[1] = new int[] { 2, 3 };

        // Function Call
        int result = FindStrongestProgrammer(N, M, power);

        if (result == -1)
        {
            Console.WriteLine(-1);
        }
        else
        {
            Console.WriteLine(result);
        }
    }
}

// this code is contributed by uttamdp_10

// JavaScript code for the above approach:

// Function to find the strongest programmer
function findStrongestProgrammer(N, M, power) {
    const deg = new Array(N + 1).fill(0);

    // Taking indegree
    for (let i = 0; i < M; i++) {
        const a = power[i][0];
        const b = power[i][1];
        deg[b]++;
    }

    let strongestProgrammer = -1;

    // Finding the strongest programmer
    for (let i = 1; i <= N; i++) {
        if (deg[i] === 0) {
            if (strongestProgrammer !== -1) {
                return -1;
            } else {
                strongestProgrammer = i;
            }
        }
    }
    
    return strongestProgrammer;
}
// Drivers code

const N = 3;
const M = 2;
const power = [[1, 2], [2, 3]];

// Function Call
const result = findStrongestProgrammer(N, M, power);

if (result === -1) {
    console.log(-1);
} else {
    console.log(result);
}

1

Time Complexity: O(N+M)
Auxiliary Space: O(N)