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:

2 1 4 3

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