Minimum index i such that all the elements from index i to given index are equal
Given an array arr[] of integers and an integer pos, the task is to find the minimum index i such that all the elements from index i to index pos are equal.
Examples:
Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 3
Output: 1
Elements in index range [1, 3] are all equal to 1.Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 5
Output: 5
Approach: Starting from index pos – 1, traverse the array in reverse and for the first index i such that arr[i] != arr[pos] print i + 1 which is the required index.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the minimum required index int minIndex(int arr[], int n, int pos) { int num = arr[pos]; // Start from arr[pos - 1] int i = pos - 1; while (i >= 0) { if (arr[i] != num) break; i--; } // All elements are equal // from arr[i + 1] to arr[pos] return i + 1; } // Driver code int main() { int arr[] = { 2, 1, 1, 1, 5, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int pos = 4; cout << minIndex(arr, n, pos); return 0; } |
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Java
// Java implementation of the approach class GFG { // Function to return the minimum required index static int minIndex(int arr[], int n, int pos) { int num = arr[pos]; // Start from arr[pos - 1] int i = pos - 1; while (i >= 0) { if (arr[i] != num) break; i--; } // All elements are equal // from arr[i + 1] to arr[pos] return i + 1; } // Driver code public static void main(String[] args) { int arr[] = { 2, 1, 1, 1, 5, 2 }; int n = arr.length; int pos = 4; System.out.println(minIndex(arr, n, pos)); } } // This code is contributed by Code_Mech. |
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Python3
# Python3 implementation of the approach # Function to return the minimum # required index def minIndex(arr, n, pos): num = arr[pos] # Start from arr[pos - 1] i = pos - 1 while (i >= 0): if (arr[i] != num): break i -= 1 # All elements are equal # from arr[i + 1] to arr[pos] return i + 1 # Driver code arr = [2, 1, 1, 1, 5, 2 ] n = len(arr) pos = 4 print(minIndex(arr, n, pos)) # This code is contributed by # Mohit Kumar 29 |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the minimum required index static int minIndex(int []arr, int n, int pos) { int num = arr[pos]; // Start from arr[pos - 1] int i = pos - 1; while (i >= 0) { if (arr[i] != num) break; i--; } // All elements are equal // from arr[i + 1] to arr[pos] return i + 1; } // Driver code public static void Main() { int []arr = { 2, 1, 1, 1, 5, 2 }; int n = arr.Length; int pos = 4; Console.WriteLine(minIndex(arr, n, pos)); } } // This code is contributed // by Akanksha Rai |
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PHP
<?php // PHP implementation of the approach // Function to return the minimum // required index function minIndex($arr, $n, $pos) { $num = $arr[$pos]; // Start from arr[pos - 1] $i = $pos - 1; while ($i >= 0) { if ($arr[$i] != $num) break; $i--; } // All elements are equal // from arr[i + 1] to arr[pos] return $i + 1; } // Driver code $arr = array(2, 1, 1, 1, 5, 2 ); $n = sizeof($arr); $pos = 4; echo minIndex($arr, $n, $pos); // This code is contributed by Ryuga ?> |
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Output:
4
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