Count pair of indices in Array having equal Prefix-MEX and Suffix-MEX
Last Updated :
23 Jul, 2025
Given an array arr[] of N elements, the task is to find the number of pairs of indices such that the Prefix-MEX of the one index is equal to the Suffix-MEX of the other index. Order of indices in pair matters.
MEX of an array refers to the smallest missing non-negative integer of the array. For the given problems (i, j) and (j, i) are considered different pairs.
Examples:
Input: A[] = {1, 0, 2, 0, 1}
Output: 11
Explanation: In array A, Prefix-MEX for each element are as {0, 2, 3, 3, 3}, similarly, Suffix-MEX for each element are {3, 3, 3, 2, 0} and all possible pairs of indexes such that prefix-MEX of one is equal to Suffix-MEX of other are (1-based indexing):
(1, 5): prefix of 1 is equal to the suffix of 5
(2, 4): prefix of 2 is equal to the suffix of 4
(3, 3): prefix of 3 is equal to the suffix of 3
(3, 2): prefix of 3 is equal to the suffix of 2
(3, 1): prefix of 3 is equal to the suffix of 1
(4, 3): prefix of 4 is equal to the suffix of 3
(4, 2): prefix of 4 is equal to the suffix of 2
(4, 1): prefix of 4 is equal to the suffix of 1
(5, 3): prefix of 5 is equal to the suffix of 3
(5, 2): prefix of 5 is equal to the suffix of 2
(5, 1): prefix of 5 is equal to the suffix of 1
Input: A[] = {1, 2, 3}
Output: 9
Naive Approach:
The most straightforward way will be to find each element's Prefix-MEX and suffix-MEX and count a total number of possible pairs and increase the count. And for each element, we have to traverse the complete array again and again therefore complexity will be of order N2.
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Efficient Approach: In this approach, we will pre-compute and store Prefix-MEX and Suffix-MEX for each element.
Follow the steps below to calculate the Prefix-MEX array for the given array:
- Find the maximum element in the given array arr.
- Create a set of integers and store the numbers from 0 to the max element in the set.
- Traverse through the array from i = 0 to N-1:
- For each element, erase that element from the set.
- Now find the smallest element remaining in the set.
- Store the value of this smallest element remaining in the set in the resultant array.
- Return the resultant array as the required answer.
Follow the steps below to implement the idea:
- Create a variable count and initialize it to 0 to store the number of indexes.
- Calculate a Prefix-MEX for arr as P for a given array.
- Reverse the given array arr and calculate the Prefix-MEX array for this revered array as S and reverse S.
- Create a map mp to count the frequency of each element in P.
- Traverse through P increment frequency of current element.
- Traverse through S and check if the current element exists in mp.
- if the current element does not exist in mp then continue.
- else add the frequency of the current element in the variable count.
- print the value of count as a required result.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the prefix MEX
// for each array element taking vector
// and its size as parameter
vector<int> Prefix_MEX(vector<int>& A, int n)
{
// Maximum element in vector A
int mx_element = *max_element(A.begin(), A.end());
// Set to store all elements for
// 0 to mx_element
set<int> s;
// Vector to store Prefix-MEX for
// given input array
vector<int> B(n);
// Store all number from 0
// to maximum element + 1 in a set
for (int i = 0; i <= mx_element + 1; i++) {
// inserting elements in set
s.insert(i);
}
// Loop to calculate MEX for each
// index in the array
for (int i = 0; i < n; i++) {
// Checking if A[i] is present
// in set
auto it = s.find(A[i]);
// If present then we erase
// that element
if (it != s.end())
s.erase(it);
// Store the first element of set
// in vector B as Mex of
// prefix vector
//cout<<*s.begin()<<" ";
B[i] = *s.begin();
}
// Return the vector B
return B;
}
void countPairs(vector<int>& arr, int N)
{
// Count variable to store number of
// indices whose prefix-mex and
// suffix-mex are equal
int count = 0;
// Vector P stores the Prefix-MEX
// for given array
vector<int> P = Prefix_MEX(arr, N);
// Reversing given array
reverse(arr.begin(), arr.end());
// vector S stores the reverse of
// Suffix-MEX for given array
vector<int> S = Prefix_MEX(arr, N);
// Reversing vector S will give
// suffix-MEX for given array
reverse(S.begin(), S.end());
// map to count frequency of each
// element in arr1
map<int, int> mp;
// Counting frequency of each
// element of arr1
for (int i = 0; i < N; i++) {
mp[P[i]]++;
}
// Traversal through arr S
for (int i = 0; i < N; i++) {
// Check if element does not exist
// in map
if (mp.find(S[i]) == mp.end()) {
continue;
}
// if exist add it's frequency
// to count
else {
count += mp[S[i]];
}
}
cout << count << endl;
}
// Driver code
int main()
{
vector<int> arr = { 1, 0, 2, 0, 1 };
int N = arr.size();
// Function Call
countPairs(arr, N);
arr = { 1, 2, 3 };
N = arr.size();
// Function Call
countPairs(arr, N);
return 0;
}
import java.util.*;
public class Main {
static List<Integer> prefixMex(List<Integer> A) {
int mxElement = Collections.max(A);
Set<Integer> s = new HashSet<>();
for (int i = 0; i < mxElement + 2; i++) {
s.add(i);
}
List<Integer> B = new ArrayList<>(Collections.nCopies(A.size(), 0));
for (int i = 0; i < A.size(); i++) {
s.remove(A.get(i));
B.set(i, Collections.min(s));
}
return B;
}
static int countPairs(List<Integer> arr) {
int n = arr.size();
List<Integer> P = prefixMex(arr);
List<Integer> S = new ArrayList<>(arr);
Collections.reverse(S);
S = prefixMex(S);
Collections.reverse(S);
Map<Integer, Integer> mp = new HashMap<>();
for (int p : P) {
mp.put(p, mp.getOrDefault(p, 0) + 1);
}
int count = 0;
for (int s : S) {
count += mp.getOrDefault(s, 0);
}
return count;
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(1, 0, 2, 0, 1);
System.out.println(countPairs(arr));
arr = Arrays.asList(1, 2, 3);
System.out.println(countPairs(arr));
}
}
from typing import List, Tuple
def prefix_mex(A: List[int]) -> List[int]:
mx_element = max(A)
s = set(range(mx_element + 2))
B = [0] * len(A)
for i, a in enumerate(A):
s.discard(a)
B[i] = min(s)
return B
def count_pairs(arr: List[int]) -> int:
n = len(arr)
P = prefix_mex(arr)
S = prefix_mex(arr[::-1])[::-1]
mp = {}
for p in P:
mp[p] = mp.get(p, 0) + 1
count = 0
for s in S:
count += mp.get(s, 0)
return count
arr = [1, 0, 2, 0, 1]
print(count_pairs(arr))
arr = [1, 2, 3]
print(count_pairs(arr))
// C# code to implement the approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// Function to find the prefix MEX
// for each array element taking vector
// and its size as parameter
static List<int> Prefix_MEX(List<int> A, int n)
{
// Maximum element in List A
int mxElement = A.Max();
// Set to store all elements for
// 0 to mx_element
HashSet<int> s = new HashSet<int>();
// Store all number from 0
// to maximum element + 1 in a set
for (int i = 0; i < mxElement + 2; i++) {
// inserting elements in set
s.Add(i);
}
// List to store Prefix-MEX for
// given input array
List<int> B = new List<int>(new int[A.Count]);
// Loop to calculate MEX for each
// index in the array
for (int i = 0; i < n; i++) {
// Checking if A[i] is present in set
// If present then we remove that element
s.Remove(A[i]);
// Store the first element of set
// in vector B as Mex of
B[i] = s.Min();
}
return B;
}
static int CountPairs(List<int> arr, int n)
{
// Vector P stores the Prefix-MEX
// for given array
List<int> P = Prefix_MEX(arr, n);
// vector S stores the reverse of
// Suffix-MEX for given array
List<int> S = new List<int>(arr);
// Reversing given array
S.Reverse();
S = Prefix_MEX(S, n);
// Reversing vector S will give
// suffix-MEX for given array
S.Reverse();
// map to count frequency of each
// element in arr1
Dictionary<int, int> mp
= new Dictionary<int, int>();
// Counting frequency of each
// element of arr1
foreach(int p in P) {
if (mp.ContainsKey(p)) {
mp[p]++;
}
else {
mp[p] = 1;
}
}
// Count variable to store number of
// indices whose prefix-mex and
// suffix-mex are equal
int count = 0;
// Traversal through arr S
foreach(int s in S) {
if (mp.ContainsKey(s)) {
count += mp[s];
}
}
return count;
}
// Driver code
public static void Main(string[] args)
{
List<int> arr = new List<int>{ 1, 0, 2, 0, 1 };
int N = arr.Count;
// Function Call
Console.WriteLine(CountPairs(arr, N));
arr = new List<int>{ 1, 2, 3 };
N = arr.Count;
// Function Call
Console.WriteLine(CountPairs(arr, N));
}
}
// This Code is Contributed by Prasad Kandekar(prasad264)
// javascript code to implement the approach
// Function to find the prefix MEX
// for each array element taking vector
// and its size as parameter
function Prefix_MEX(A, n)
{
// Maximum element in vector A
let mx_element = Number.MIN_SAFE_INTEGER;
for(let i=0; i<A.length; i++)
mx_element=Math.max(mx_element, A[i]);
// Set to store all elements for
// 0 to mx_element
let s=new Set();
// Vector to store Prefix-MEX for
// given input array
let B=new Array(n);
// Store all number from 0
// to maximum element + 1 in a set
for (let i = 0; i <= mx_element + 1; i++) {
// inserting elements in set
s.add(i);
}
// Loop to calculate MEX for each
// index in the array
for (let i = 0; i < n; i++) {
// Checking if A[i] is present
// in set
//auto it = s.find(A[i]);
// If present then we erase
// that element
if (s.has(A[i]))
s.delete(A[i]);
// Store the first element of set
// in vector B as Mex of
// prefix vector
const [first]=s;
B[i] = first;
}
// Return the vector B
return B;
}
function countPairs(arr, N)
{
// Count variable to store number of
// indices whose prefix-mex and
// suffix-mex are equal
let count = 0;
// Vector P stores the Prefix-MEX
// for given array
let P = Prefix_MEX(arr, N);
// Reversing given array
arr.reverse();
// vector S stores the reverse of
// Suffix-MEX for given array
let S = Prefix_MEX(arr, N);
// Reversing vector S will give
// suffix-MEX for given array
S.reverse();
// map to count frequency of each
// element in arr1
let mp=new Map();
// Counting frequency of each
// element of arr1
for (let i = 0; i < N; i++) {
//mp[P[i]]++;
if(mp.has(P[i]))
mp.set(P[i], mp.get(P[i])+1);
else
mp.set(P[i],1);
}
// Traversal through arr S
for (let i = 0; i < N; i++) {
// Check if element does not exist
// in map
if (!mp.has(S[i])) {
continue;
}
// if exist add it's frequency
// to count
else {
count += mp.get(S[i]);
}
}
console.log(count);
}
// Driver code
let arr = [ 1, 0, 2, 0, 1 ];
let N = arr.length;
// Function Call
countPairs(arr, N);
arr = [ 1, 2, 3 ];
N = arr.length;
// Function Call
countPairs(arr, N);
// This code is contributed by poojaagarwal2.
Time Complexity: O(N * log N), log N for inserting and deleting elements from the set, and O(N) for traversing the array.
Auxiliary Space: O(N) for storing Prefix-MEX array.
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