Maximum Fixed Point (Value equal to index) in a given Array Last Updated : 18 Jul, 2021 Comments Improve Suggest changes Like Article Like Report Given an array arr[] of size N, the task is to find the maximum index i such that arr[i] is equal to i. If there is no such index in the array arr[] then print -1. Examples: Input: arr[ ] = {-10, -5, 0, 3, 7}Output: 3Explanation: Only for i=3, arr[3] = 3 Input: arr[ ] = {0, 2, 5, 8, 4}Output: 4 Approach: Follow the steps below to solve this problem: Iterate in the range [N-1, 0] using the variable i: If the current element is equal to i, then print i and return.If there is no such index, then print -1. Below is the implementation of the above approach: C++ // C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find the maximum index // i such that arr[i] is equal to i void findLargestIndex(int arr[], int n) { // Traversing the array from // backwards for (int i = n - 1; i >= 0; i--) { // If arr[i] is equal to i if (arr[i] == i) { cout << i << endl; return; } } // If there is no such index cout << -1 << endl; } // Driver code int main() { // Given Input int arr[] = { -10, -5, 0, 3, 7 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Call findLargestIndex(arr, n); return 0; } Java // Java implementation of the above approach import java.io.*; class GFG { // Function to find the maximum index // i such that arr[i] is equal to i static void findLargestIndex(int arr[], int n) { // Traversing the array from // backwards for (int i = n - 1; i >= 0; i--) { // If arr[i] is equal to i if (arr[i] == i) { System.out.println(i); return; } } // If there is no such index System.out.println(-1); } // Driver code public static void main(String[] args) { // Given Input int arr[] = { -10, -5, 0, 3, 7 }; int n = arr.length; // Function Call findLargestIndex(arr, n); } } // This code is contributed by Potta Lokesh Python # Python implementation of the above approach # Function to find the maximum index # i such that arr[i] is equal to i def findLargestIndex(arr, n): # Traversing the array from # backwards for i in range (n): # If arr[i] is equal to i if (arr[i] == i) : print( i ) return # If there is no such index print( -1 ) # Driver code # Given Input arr = [-10, -5, 0, 3, 7 ] n = len(arr) # Function Call findLargestIndex(arr, n) # This code is contributed by shivanisinghss2110 C# // C# implementation of the above approach using System; class GFG { // Function to find the maximum index // i such that arr[i] is equal to i static void findLargestIndex(int []arr, int n) { // Traversing the array from // backwards for (int i = n - 1; i >= 0; i--) { // If arr[i] is equal to i if (arr[i] == i) { Console.Write(i); return; } } // If there is no such index Console.Write(-1); } // Driver code public static void Main(String[] args) { // Given Input int []arr = { -10, -5, 0, 3, 7 }; int n = arr.Length; // Function Call findLargestIndex(arr, n); } } // This code is contributed by shivanisinghss2110 JavaScript <script> // JavaScript implementation of the above approach // Function to find the maximum index // i such that arr[i] is equal to i function findLargestIndex( arr, n) { // Traversing the array from // backwards for (var i = n - 1; i >= 0; i--) { // If arr[i] is equal to i if (arr[i] == i) { document.write(i); return ; } } // If there is no such index document.write(-1); } // Driver code // Given Input var arr = [ -10, -5, 0, 3, 7 ]; var n = arr.length; // Function Call findLargestIndex(arr, n); // This code is contributed by shivanisinghss2110 </script> Output3 Time Complexity: O(N)Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Find the peak index of a given array S siddharth_25 Follow Improve Article Tags : DSA Similar Reads For each Array index find the maximum value among all M operations Given an array arr[] of size N initially filled with 0 and another array Positions[] of size M, the task is to return the maximum value for each index after performing the following M operations: Make the value at index Positions[i] equal to 0All the numbers to the right of Positions[i] will be one 6 min read Find Maximum value of abs(i - j) * min(arr[i], arr[j]) in an array arr[] Given an array of n distinct elements. 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