Maximize the Minimum MEX

Given two integers n, m and an array arr[], where arr[i] represent a subarray from index arr[i][0] to arr[i][1]. The task is to construct an array ans[] of size n, such that the minimum MEX among the given m subarrays is as large as possible.

Note: There may exist multiple ans[], return one of them.

Examples:

Input: n=5, m=3, arr = [ [1,3], [2,5], [4,5] ]
Output: [1, 0, 2, 1, 0]
Explanation: MEX of first subarray ([1, 0, 2]) is 3, MEX of second subarray ([0, 2, 1, 0]) is 3 and MEX of third subarray ([0,1]) is 2. Hence, the minimum MEX is 2 which is the maximum possible.

Input: n=4, m=2, arr=[ [1,4], [2,4] ]
Output: [3, 0, 1, 2]
Explanation: MEX of first subarray ([3, 0, 1, 2]) is 4, and MEX of second subarray ([1, 2, 0]) is 3 . Hence, the minimum MEX is 3 which is the maximum possible.

Approach:

It can be observed that the answer cannot be greater than the size of the smallest subarrays among the given subarrays because the MEX of the array of size n cannot be greater than. Let ans denote the size of the smallest subarray. So we can simple construct the array ans [0, 1, 2, .... ans-1, 0, 1, .... ]. Now since each of the given subarray has least ans size, so each of the subrarray will contain all the numbers from 0 to ans-1.

Follow the steps to solve the given problem:

• Iterate over m subarrays.
• Determine the minimum size subarray and stores its size in a variable.
• Run a loop from i is equal to 0 to n-1, and output print (i % ans), for each i.

Below is the implementation of above approach:

#include <bits/stdc++.h>

using namespace std;

void solve(int n, int m, vector<vector<int> >& subarrays)
{
    // Initialize ans with the maximum possible value of n.
    int ans = n;

    // Iterate over the subarrays to find the minimum size
    // among them.
    for (int i = 0; i < m; i++) {
        int x = subarrays[i][0];
        int y = subarrays[i][1];
        int diff = y - x;

        // Update ans with the minimum size among subarrays.
        ans = min(ans, diff + 1);
    }
  
   // Construct the array to maximize the minimum MEX.
    for (int i = 0; i < n; i++) {
        cout << (i % ans) << " "; 
    }
}

// Driver Code

int main()
{
    int n=4, m=2;
    vector<vector<int>> subarrays={{1,4}, {2,4}} ;
      solve(n,m,subarrays);
   
}

// Java Implmentation
import java.util.ArrayList;
import java.util.List;

public class Main {
    public static void main(String[] args) {
        int n = 4;
        int m = 2;
        List<List<Integer>> subarrays = new ArrayList<>();
        subarrays.add(List.of(1, 4));
        subarrays.add(List.of(2, 4));
        solve(n, m, subarrays);
    }

    public static void solve(int n, int m, List<List<Integer>> subarrays) {
        // Initialize ans with the maximum possible value of n.
        int ans = n;

        // Iterate over the subarrays to find the minimum size among them.
        for (List<Integer> subarray : subarrays) {
            int x = subarray.get(0);
            int y = subarray.get(1);
            int diff = y - x;

            // Update ans with the minimum size among subarrays.
            ans = Math.min(ans, diff + 1);
        }

        // Construct the array to maximize the minimum MEX.
        for (int i = 0; i < n; i++) {
            System.out.print(i % ans + " ");
        }
    }
}
// This code is contributed by Sakshi

def construct_array(n, m, subarrays):
    ans = n

    for i in range(m):
        x, y = subarrays[i]
        diff = y - x + 1
        ans = min(ans, diff)

    result = []
    for i in range(n):
        result.append(i % ans)

    return result

# Driver Code
if __name__ == "__main__":
    n = 5
    m = 3
    subarrays = [[1, 3], [2, 5], [4, 5]]
    ans = construct_array(n, m, subarrays)
    print(ans)

    

// C# Implmentation

using System;
using System.Collections.Generic;

class GFG {
    public static void Main()
    {
        int n = 4;
        int m = 2;
        List<List<int>> subarrays = new List<List<int>>();
        subarrays.Add(new List<int>() { 1, 4 });
        subarrays.Add(new List<int>() { 2, 4 });
        solve(n, m, subarrays);
    }

    static void solve(int n, int m,
                      List<List<int> > subarrays)
    {
        // Initialize ans with the maximum possible
        // value of n.
        int ans = n;

        // Iterate over the subarrays to find the minimum
        // size among them.
        foreach(var subarray in subarrays)
        {
            int x = subarray[0];
            int y = subarray[1];
            int diff = y - x;

            // Update ans with the minimum size among
            // subarrays.
            ans = Math.Min(ans, diff + 1);
        }

        // Construct the array to maximize the minimum MEX.
        for (int i = 0; i < n; i++) {
            Console.Write((i % ans) + " ");
        }
    }
}

// This code is contributed by ragul21

function solve(n, m, subarrays) {
    // Initialize ans with the maximum possible value of n.
    let ans = n;

    // Iterate over the subarrays to find the minimum size
    // among them.
    for (let i = 0; i < m; i++) {
        let x = subarrays[i][0];
        let y = subarrays[i][1];
        let diff = y - x;

        // Update ans with the minimum size among subarrays.
        ans = Math.min(ans, diff + 1);
    }

    // Construct the array to maximize the minimum MEX.
    for (let i = 0; i < n; i++) {
        process.stdout.write((i % ans) + " ");
    }
}

// Driver Code
function main() {
    let n = 4, m = 2;
    let subarrays = [[1, 4], [2, 4]];
    solve(n, m, subarrays);
}

main();

0 1 0 1 0 

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