Construct MEX array from the given array

Given an array arr[] having N distinct positive elements, the task is to generate another array B[] such that, for every i^th index in the array, arr[], B[i] is the minimum positive number missing from arr[] excluding arr[i].

Examples:

Input: arr[] = {2, 1, 5, 3}
Output: B[] = {2, 1, 4, 3} 
Explanation: After excluding the arr[0], the array is {1, 5, 3}, and the minimum positive number which is not present in this array is 2. Therefore, B[0] = 2. Similarly, after excluding arr[1], arr[2], arr[3], the minimum positive numbers which are not present in the array are 1, 4 and 3, respectively. Hence, B[1] = 1, B[2] = 4, B[3] = 3.

Input: arr[] = {1, 9, 2, 4}
Output: B[] = {1, 3, 2, 3}

Naive Approach: The simplest approach to solve this problem is to traverse the array arr[] and for every index i, initialize an array hash[] and for every index j ( where j ? i), update hash[arr[j]] =1. Now traverse array hash[] from index 1 and find the minimum index k for which hash[k] = 0 and update B[i] = k. Finally, print the array B[] after completing the above step.

Time Complexity: O(N²) where N is the length of the given array.
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to calculate MEX of the array arr[] and traverse the array arr[]. If arr[i] is less than MEX of the array arr[] then MEX excluding this element will be arr[i] itself, and if arr[i] is greater than MEX of array A[] then MEX of the array will not change after excluding this element.

Follow the steps below to solve the problem:

1. Initialize an array, say hash[], to store whether the value i is present in the array arr[] or not. If i is present hash[i] = 1 else hash[i] = 0.
2. Initialize a variable MexOfArr to store MEX of array arr[] and traverse array hash[] from 1 to find the minimum index j for which hash[j] = 0, which implies that the value j is not present in the array arr[] and store MexOfArr = j.
3. Now traverse the array, arr[] and if arr[i] is less than MexOfArr, then store B[i] = arr[i] else B[i] = MexOfArr.
4. After completing the above steps, print elements of the array B[] as the required answer.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

#define MAXN 100001

// Function to construct array B[] that
// stores MEX of array A[] excluding A[i]
void constructMEX(int arr[], int N)
{
    // Stores elements present in arr[]
    int hash[MAXN] = { 0 };

    // Mark all values 1, if present
    for (int i = 0; i < N; i++) {

        hash[arr[i]] = 1;
    }

    // Initialize variable to store MEX
    int MexOfArr;

    // Find MEX of arr[]
    for (int i = 1; i < MAXN; i++) {
        if (hash[i] == 0) {
            MexOfArr = i;
            break;
        }
    }

    // Stores MEX for all indices
    int B[N];

    // Traverse the given array
    for (int i = 0; i < N; i++) {

        // Update MEX
        if (arr[i] < MexOfArr)
            B[i] = arr[i];

        // MEX default
        else
            B[i] = MexOfArr;
    }

    // Print the array B
    for (int i = 0; i < N; i++)
        cout << B[i] << ' ';
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 1, 5, 3 };

    // Given size
    int N = sizeof(arr)
            / sizeof(arr[0]);

    // Function call
    constructMEX(arr, N);
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG{
    
static int MAXN = 100001;
  
// Function to construct array
// B[] that stores MEX of array 
// A[] excluding A[i]
static void constructMEX(int arr[], 
                         int N)
{
  // Stores elements present 
  // in arr[]
  int hash[] = new int[MAXN];
  for (int i = 0; i < N; i++) 
  {
    hash[i] = 0;
  }

  // Mark all values 1, if
  // present
  for (int i = 0; i < N; i++) 
  {
    hash[arr[i]] = 1;
  }

  // Initialize variable to 
  // store MEX
  int MexOfArr = 0 ;

  // Find MEX of arr[]
  for (int i = 1; i < MAXN; i++)
  {
    if (hash[i] == 0) 
    {
      MexOfArr = i;
      break;
    }
  }

  // Stores MEX for all 
  // indices
  int B[] = new int[N];

  // Traverse the given array
  for (int i = 0; i < N; i++) 
  {
    // Update MEX
    if (arr[i] < MexOfArr)
      B[i] = arr[i];

    // MEX default
    else
      B[i] = MexOfArr;
  }

  // Print the array B
  for (int i = 0; i < N; i++)
    System.out.print(B[i] + " ");
}

// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {2, 1, 5, 3};

  // Size of array
  int N = arr.length;

  // Function call
  constructMEX(arr, N);
}
}

// This code is contributed by sanjoy_62

# Python3 program for the 
# above approach

MAXN = 100001

# Function to construct 
# array B[] that stores 
# MEX of array A[] excluding 
# A[i]
def constructMEX(arr, N):
  
    # Stores elements present 
    # in arr[]
    hash = [0] * MAXN

    # Mark all values 1, 
    # if present
    for i in range(N):
        hash[arr[i]] = 1

    # Initialize variable to 
    # store MEX
    MexOfArr = 0

    # Find MEX of arr[]
    for i in range(1, MAXN):
        if (hash[i] == 0):
            MexOfArr = i
            break

    # Stores MEX for all
    # indices
    B = [0] * N

    # Traverse the given array
    for i in range(N):

        # Update MEX
        if (arr[i] < MexOfArr):
            B[i] = arr[i]

        # MEX default
        else:
            B[i] = MexOfArr

    # Print array B
    for i in range(N):
        print(B[i], end = " ")

# Driver Code
if __name__ == '__main__':
  
    # Given array
    arr = [2, 1, 5, 3]

    # Given size
    N = len(arr)

    # Function call
    constructMEX(arr, N)

# This code is contributed by Mohit Kumar 29

// C# program for the above approach
using System;

class GFG{

static int MAXN = 100001;
  
// Function to construct array
// B[] that stores MEX of array 
// A[] excluding A[i]
static void constructMEX(int[] arr, 
                         int N)
{
  
  // Stores elements present 
  // in arr[]
  int[] hash = new int[MAXN];
  for(int i = 0; i < N; i++) 
  {
    hash[i] = 0;
  }

  // Mark all values 1, if
  // present
  for(int i = 0; i < N; i++) 
  {
    hash[arr[i]] = 1;
  }

  // Initialize variable to 
  // store MEX
  int MexOfArr = 0;

  // Find MEX of arr[]
  for(int i = 1; i < MAXN; i++)
  {
    if (hash[i] == 0) 
    {
      MexOfArr = i;
      break;
    }
  }
  
  // Stores MEX for all 
  // indices
  int[] B = new int[N];

  // Traverse the given array
  for(int i = 0; i < N; i++) 
  {
    
    // Update MEX
    if (arr[i] < MexOfArr)
      B[i] = arr[i];

    // MEX default
    else
      B[i] = MexOfArr;
  }

  // Print the array B
  for(int i = 0; i < N; i++)
    Console.Write(B[i] + " ");
}

// Driver Code
public static void Main() 
{
  
  // Given array arr[]
  int[] arr = { 2, 1, 5, 3 };
  
  // Size of array
  int N = arr.Length;
  
  // Function call
  constructMEX(arr, N);
}
}

// This code is contributed by code_hunt

<script>
// Javascript program for the above approach

var MAXN = 100001;

// Function to construct array B[] that
// stores MEX of array A[] excluding A[i]
function constructMEX(arr, N)
{
    // Stores elements present in arr[]
    var hash = Array(MAXN).fill(0);

    // Mark all values 1, if present
    for (var i = 0; i < N; i++) {

        hash[arr[i]] = 1;
    }

    // Initialize variable to store MEX
    var MexOfArr;

    // Find MEX of arr[]
    for (var i = 1; i < MAXN; i++) {
        if (hash[i] == 0) {
            MexOfArr = i;
            break;
        }
    }

    // Stores MEX for all indices
    var B = Array(N);

    // Traverse the given array
    for (var i = 0; i < N; i++) {

        // Update MEX
        if (arr[i] < MexOfArr)
            B[i] = arr[i];

        // MEX default
        else
            B[i] = MexOfArr;
    }

    // Print the array B
    for (var i = 0; i < N; i++)
        document.write( B[i] + ' ');
}

// Driver Code
// Given array
var arr = [2, 1, 5, 3];
// Given size
var N = arr.length;
// Function call
constructMEX(arr, N);

</script> 

2 1 4 3

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