Construct MEX array from the given array
Last Updated :
13 Jan, 2022
Given an array arr[] having N distinct positive elements, the task is to generate another array B[] such that, for every ith 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:
- 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.
- 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.
- Now traverse the array, arr[] and if arr[i] is less than MexOfArr, then store B[i] = arr[i] else B[i] = MexOfArr.
- After completing the above steps, print elements of the array B[] as the required answer.
Below is the implementation of the above approach:
C++
// 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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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