Given an array A[]. The task is to determine if it is possible to choose two indices āiā and ājā such that the below conditions gets satisfied:-
- A[i] is not equal to A[j].
- A[A[i]] is equal to A[A[j]].
Note: The value of the elements in an array is less than the value of N i.e. For every i, arr[i] < N.Ā
Examples:Ā
Input: N = 4, A[] = {1, 1, 2, 3}
Output: Yes
As A[3] != to A[1] but A[A[3]] == A[A[1]]
Input: N = 4, A[] = {2, 1, 3, 3}
Output: No
As A[A[3]] == A[A[4]] but A[3] == A[4]
Approach:Ā Ā
- Start traversing the Array Arr[] by running two loops.
- The variable i point at the index 0 and variable j point to the next of i.
- If Arr[i] is not equal to Arr[j] then check if Arr[Arr[i] ā 1] is equal to Arr[Arr[j] ā 1]. If yes then return true.Ā
Else check Arr[Arr[i]- 1] and Arr[Arr[j] ā 1] for other indices also. - Repeat the above step till all the elements/index gets traversed.
- If no such indices found return false.
Below is the implementation of the above approach:Ā
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;Ā
// Function that will tell whether// such Indices present or Not.bool checkIndices(int Arr[], int N){Ā
Ā Ā Ā Ā for (int i = 0; i < N - 1; i++) {Ā Ā Ā Ā Ā Ā Ā Ā for (int j = i + 1; j < N; j++) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 1st condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to Arr[j] or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[i] != Arr[j]) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 2nd condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā return false;}Ā
// Driver Codeint main(){Ā Ā Ā Ā int Arr[] = { 3, 2, 1, 1, 4 };Ā Ā Ā Ā int N = sizeof(Arr) / sizeof(Arr[0]);Ā
Ā Ā Ā Ā // Calling function.Ā Ā Ā Ā checkIndices(Arr, N) ? cout << "Yes"Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā : cout << "No";Ā
Ā Ā Ā Ā return 0;} |
Java
// Java implementation of the above approach Ā
// Function that calculates marks. class GFG{ Ā Ā Ā Ā static boolean checkIndices(int Arr[], int N) Ā Ā Ā Ā { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < N - 1; i++) { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā for (int j = i + 1; j < N; j++) { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 1st condition i.e whether Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to Arr[j] or not Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[i] != Arr[j]) { Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 2nd condition i.e whether Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j]] or not. Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā } Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā } Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā return false;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Driver code Ā Ā Ā Ā public static void main(String args[]) Ā Ā Ā Ā { Ā Ā Ā Ā Ā Ā Ā Ā int Arr[] = { 3, 2, 1, 1, 4 }; Ā Ā Ā Ā Ā Ā Ā Ā int N = Arr.length;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if(checkIndices(Arr, N))Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("Yes");Ā Ā Ā Ā Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("No");Ā Ā Ā Ā } } Ā
// This code is Contributed by // Naman_Garg |
Python 3
# Python 3 implementation of the # above approachĀ
# Function that will tell whether# such Indices present or Not.def checkIndices(Arr, N):Ā
Ā Ā Ā Ā for i in range(N - 1):Ā Ā Ā Ā Ā Ā Ā Ā for j in range(i + 1, N):Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Checking 1st condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Arr[i] equal to Arr[j] or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[i] != Arr[j]):Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Checking 2nd condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Arr[Arr[i]] equal to Arr[Arr[j]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]):Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return TrueĀ
Ā Ā Ā Ā return FalseĀ
# Driver Codeif __name__ == "__main__":Ā Ā Ā Ā Ā Ā Ā Ā Ā Arr = [ 3, 2, 1, 1, 4 ]Ā Ā Ā Ā N =len(Arr)Ā
Ā Ā Ā Ā # Calling function.Ā Ā Ā Ā if checkIndices(Arr, N):Ā Ā Ā Ā Ā Ā Ā Ā print("Yes")Ā Ā Ā Ā Ā else:Ā Ā Ā Ā Ā Ā Ā Ā print("No")Ā
# This code is contributed by ita_c |
C#
// C# implementation of the above approach using System;Ā
class GFG{Ā Ā Ā Ā Ā // Function that calculates marks. static bool checkIndices(int []Arr, int N) { Ā Ā Ā Ā for (int i = 0; i < N - 1; i++)Ā Ā Ā Ā { Ā Ā Ā Ā Ā Ā Ā Ā for (int j = i + 1; j < N; j++) Ā Ā Ā Ā Ā Ā Ā Ā { Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 1st condition i.e whether Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to Arr[j] or not Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[i] != Arr[j]) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā { Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 2nd condition i.e Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // whether Arr[Arr[i]] equalĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // to Arr[Arr[j]] or not. Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true; Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā } Ā Ā Ā Ā Ā Ā Ā Ā } Ā Ā Ā Ā } Ā Ā Ā Ā return false; } Ā
// Driver code static public void Main (){Ā Ā Ā Ā int []Arr = { 3, 2, 1, 1, 4 }; Ā Ā Ā Ā int N = Arr.Length; Ā Ā Ā Ā Ā Ā Ā Ā Ā if(checkIndices(Arr, N)) Ā Ā Ā Ā Ā Ā Ā Ā Console.WriteLine("Yes"); Ā Ā Ā Ā elseĀ Ā Ā Ā Ā Ā Ā Ā Console.WriteLine("No"); } } Ā
// This code is Contributed by Sachin |
PHP
<?php// PHP implementation of the // above approachĀ
// Function that will tell whether// such Indices present or Not.function checkIndices($Arr, $N){Ā Ā Ā Ā for ($i = 0; $i < $N - 1; $i++) Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā for ($j = $i + 1;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā $j < $N; $j++)Ā Ā Ā Ā Ā Ā Ā Ā {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 1st condition i.e Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // whether Arr[i] equal toĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[j] or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if ($Arr[$i] != $Arr[$j]) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 2nd condition i.e Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // whether Arr[Arr[i]] equal to Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[j]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if ($Arr[$Arr[$i] - 1] == $Arr[$Arr[$j] - 1])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā return false;}Ā
// Driver Code$Arr = array(3, 2, 1, 1, 4);$N = sizeof($Arr);Ā
// Calling function.if(checkIndices($Arr, $N)) Ā Ā Ā Ā echo "Yes";elseĀ Ā Ā Ā echo "No";Ā
// This code is contributed // by Akanksha Rai?> |
Javascript
<script>Ā
// Javascript implementation of the above approachĀ
// Function that will tell whether// such Indices present or Not.function checkIndices(Arr, N){Ā
Ā Ā Ā Ā for (var i = 0; i < N - 1; i++) {Ā Ā Ā Ā Ā Ā Ā Ā for (var j = i + 1; j < N; j++) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 1st condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to Arr[j] or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[i] != Arr[j]) {Ā
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking 2nd condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }Ā
Ā Ā Ā Ā return false;}Ā
// Driver Codevar Arr = [ 3, 2, 1, 1, 4 ];var N = Arr.length;Ā
// Calling function.checkIndices(Arr, N) ? document.write( "Yes")Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā : document.write( "No");Ā
Ā
</script> |
Output
Yes
Complexity Analysis:
- Time Complexity: O(N2)
- Auxiliary Space: O(1)
Optimization: The above approach used can be optimized to O(n) with the help of an unordered_map.
The steps used in this approach are as follows:
- First, we use an unordered_map to store the mapping of Arr[i] to i+1.Ā
- Then, we iterate through the array and check if Arr[i] is already present in the hash table(unordered_map).Ā
- If it is present, we check if the corresponding index in the hash table has the same value as Arr[i]. If yes then we have found the required indices and return ātrueā. If not, we continue the iteration.Ā
- If we reach the end of the iteration and have not found the required indices, we return false.
Below is the code for the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;Ā
// Function that will tell whether// such Indices present or Not.bool checkIndices(int Arr[], int N){Ā Ā Ā Ā // Hash table to store the mapping of Arr[i] to i+1Ā Ā Ā Ā unordered_map<int, int> mp;Ā
Ā Ā Ā Ā for (int i = 0; i < N; i++) {Ā Ā Ā Ā Ā Ā Ā Ā // 1st condition is checked here i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to any other previous elementĀ Ā Ā Ā Ā Ā Ā Ā // or notĀ Ā Ā Ā Ā Ā Ā Ā if (mp.find(Arr[i]) != mp.end()) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int j = mp[Arr[i]];Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking the second condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j-1]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Here we are taking j-1 as we are storing Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // the Arr[i] in i+1.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[j-1] == Arr[i]) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā mp[Arr[i]] = i+1;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā return false;}Ā
// Driver Codeint main(){Ā Ā Ā Ā int Arr[] = { 3, 2, 1, 1, 4 };Ā Ā Ā Ā int N = sizeof(Arr) / sizeof(Arr[0]);Ā
Ā Ā Ā Ā // Calling function.Ā Ā Ā Ā checkIndices(Arr, N) ? cout << "Yes"Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā : cout << "No";Ā
Ā Ā Ā Ā return 0;}Ā
// This code is contributed by Pushpesh Raj. |
Python3
# Python 3 implementation of the above approachĀ
# Function that will tell whether# such Indices present or Not.def checkIndices(arr, n):Ā Ā Ā Ā # Dictionary to store the mapping of arr[i] to i+1Ā Ā Ā Ā mp = {}Ā
Ā Ā Ā Ā for i in range(n):Ā Ā Ā Ā Ā Ā Ā Ā # 1st condition is checked here i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā # arr[i] equal to any other previous elementĀ Ā Ā Ā Ā Ā Ā Ā # or notĀ Ā Ā Ā Ā Ā Ā Ā if arr[i] in mp:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā j = mp[arr[i]]Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Checking the second condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # arr[arr[i]] equal to arr[arr[j-1]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # Here we are taking j-1 as we are storingĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā # the arr[i] in i+1.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if arr[j-1] == arr[i]:Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return TrueĀ Ā Ā Ā Ā Ā Ā Ā mp[arr[i]] = i+1Ā
Ā Ā Ā Ā return FalseĀ
# Driver Codeif __name__ == "__main__":Ā Ā Ā Ā arr = [3, 2, 1, 1, 4]Ā Ā Ā Ā n = len(arr)Ā
Ā Ā Ā Ā # Calling function.Ā Ā Ā Ā print("Yes" if checkIndices(arr, n) else "No") |
Java
// Java implementation of the above approachimport java.io.*;Ā
class GFG {Ā Ā Ā Ā // Function that will tell whetherĀ Ā Ā Ā // such Indices present or Not.Ā Ā Ā Ā static boolean checkIndices(int Arr[], int N)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // Hash table to store the mapping of Arr[i] to i+1Ā Ā Ā Ā Ā Ā Ā Ā java.util.Map<Integer, Integer> mpĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = new java.util.HashMap<>();Ā
Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < N; i++) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // 1st condition is checked here i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to any other previous elementĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (mp.containsKey(Arr[i])) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int j = mp.get(Arr[i]);Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking the second condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j-1]] orĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // not. Here we are taking j-1 as we areĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // storing the Arr[i] in i+1.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[j - 1] == Arr[i]) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā mp.put(Arr[i], i + 1);Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā return false;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Driver CodeĀ Ā Ā Ā public static void main(String[] args)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā int Arr[] = { 3, 2, 1, 1, 4 };Ā Ā Ā Ā Ā Ā Ā Ā int N = Arr.length;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Calling function.Ā Ā Ā Ā Ā Ā Ā Ā if (checkIndices(Arr, N)) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("Yes");Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā else {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("No");Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā }} |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;Ā
class Program {Ā Ā Ā Ā // Function that will tell whetherĀ Ā Ā Ā // such Indices present or Not.Ā Ā Ā Ā static bool checkIndices(int[] Arr, int N)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // Hash table to store the mapping of Arr[i] to i+1Ā Ā Ā Ā Ā Ā Ā Ā Dictionary<int, int> mpĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = new Dictionary<int, int>();Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < N; i++) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // 1st condition is checked here i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[i] equal to any other previous elementĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // or notĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (mp.ContainsKey(Arr[i])) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int j = mp[Arr[i]];Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking the second condition i.e whetherĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Arr[Arr[i]] equal to Arr[Arr[j-1]] orĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // not. Here we are taking j-1 as we areĀ Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // storing the Arr[i] in i+1.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (Arr[j - 1] == Arr[i]) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā mp[Arr[i]] = i + 1;Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā return false;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Driver CodeĀ Ā Ā Ā static void Main(string[] args)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā int[] Arr = { 3, 2, 1, 1, 4 };Ā Ā Ā Ā Ā Ā Ā Ā int N = Arr.Length;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Calling function.Ā Ā Ā Ā Ā Ā Ā Ā Console.WriteLine(checkIndices(Arr, N) ? "Yes"Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā : "No");Ā Ā Ā Ā }} |
Javascript
// Function that will tell whether such Indices present or Not.function checkIndices(arr) {Ā Ā Ā Ā // Dictionary to store the mapping of arr[i] to i+1Ā Ā Ā Ā let mp = {};Ā
Ā Ā Ā Ā for (let i = 0; i < arr.length; i++) {Ā Ā Ā Ā Ā Ā Ā Ā // 1st condition is checked here i.e whether arr[i] equal to any other previous element or notĀ Ā Ā Ā Ā Ā Ā Ā if (arr[i] in mp) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā let j = mp[arr[i]];Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Checking the second condition i.e whether arr[arr[i]] equal to arr[arr[j-1]] or not.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā // Here we are taking j-1 as we are storing the arr[i] in i+1.Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (arr[j-1] == arr[i]) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return true;Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā }Ā Ā Ā Ā Ā Ā Ā Ā mp[arr[i]] = i+1;Ā Ā Ā Ā }Ā
Ā Ā Ā Ā return false;}Ā
// Driver Codelet arr = [3, 2, 1, 1, 4];Ā
// Calling function.console.log(checkIndices(arr) ? "Yes" : "No"); |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)
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