Given a binary array of N numbers. The task is to find the smallest index such that there are either no 1’s or 0’s to the right of the index.
Note: The array will have at least one 0 and one 1.
Examples:
Input: a[] = {1, 1, 1, 0, 0, 1, 0, 1, 1}
Output: 6
At 6th index, there are no 0’s to the right of the index.
Input: a[] = {0, 1, 0, 0, 0}
Output: 1
Approach: Store the rightmost occurring index of both 1 and 0 and return the minimum of both.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the smallest index// such that there are no 0 or 1 to its rightint smallestIndex(int a[], int n){ // Initially int right1 = 0, right0 = 0; // Traverse in the array for (int i = 0; i < n; i++) { // Check if array element is 1 if (a[i] == 1) right1 = i; // a[i] = 0 else right0 = i; } // Return minimum of both return min(right1, right0);}// Driver codeint main(){ int a[] = { 1, 1, 1, 0, 0, 1, 0, 1, 1 }; int n = sizeof(a) / sizeof(a[0]); cout << smallestIndex(a, n); return 0;} |
Java
// Java program to implement// the above approachclass GFG{ // Function to find the smallest index// such that there are no 0 or 1 to its rightstatic int smallestIndex(int []a, int n){ // Initially int right1 = 0, right0 = 0; // Traverse in the array for (int i = 0; i < n; i++) { // Check if array element is 1 if (a[i] == 1) right1 = i; // a[i] = 0 else right0 = i; } // Return minimum of both return Math.min(right1, right0);}// Driver codepublic static void main(String[] args){ int []a = { 1, 1, 1, 0, 0, 1, 0, 1, 1 }; int n = a.length; System.out.println(smallestIndex(a, n));}}// This code is contributed // by Code_Mech. |
Python3
# Python 3 program to implement# the above approach# Function to find the smallest # index such that there are no # 0 or 1 to its rightdef smallestIndex(a, n): # Initially right1 = 0 right0 = 0 # Traverse in the array for i in range(n): # Check if array element is 1 if (a[i] == 1): right1 = i # a[i] = 0 else: right0 = i # Return minimum of both return min(right1, right0)# Driver codeif __name__ == '__main__': a = [1, 1, 1, 0, 0, 1, 0, 1, 1] n = len(a) print(smallestIndex(a, n)) # This code is contributed by# Surendra_Gangwar |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to find the smallest index// such that there are no 0 or 1 to its rightstatic int smallestIndex(int []a, int n){ // Initially int right1 = 0, right0 = 0; // Traverse in the array for (int i = 0; i < n; i++) { // Check if array element is 1 if (a[i] == 1) right1 = i; // a[i] = 0 else right0 = i; } // Return minimum of both return Math.Min(right1, right0);}// Driver codepublic static void Main(){ int []a = { 1, 1, 1, 0, 0, 1, 0, 1, 1 }; int n = a.Length; Console.Write(smallestIndex(a, n));}}// This code is contributed // by Akanksha Rai |
PHP
<?php// PHP program to implement// the above approach// Function to find the smallest index// such that there are no 0 or 1 to its rightfunction smallestIndex($a, $n){ // Initially $right1 = 0; $right0 = 0; // Traverse in the array for ($i = 0; $i < $n; $i++) { // Check if array element is 1 if ($a[$i] == 1) $right1 = $i; // a[i] = 0 else $right0 = $i; } // Return minimum of both return min($right1, $right0);}// Driver code$a = array(1, 1, 1, 0, 0, 1, 0, 1, 1);$n = sizeof($a);echo smallestIndex($a, $n);// This code is contributed by Akanksha Rai?> |
Javascript
<script>// Javascript program to implement// the above approach // Function to find the smallest index // such that there are no 0 or 1 to its right function smallestIndex(a, n) { // Initially let right1 = 0, right0 = 0; // Traverse in the array let i; for (i = 0; i < n; i++) { // Check if array element is 1 if (a[i] == 1) right1 = i; // a[i] = 0 else right0 = i; } // Return minimum of both return Math.min(right1, right0); } // Driver Code var a = [ 1, 1, 1, 0, 0, 1, 0, 1, 1 ]; let n = a.length; document.write(smallestIndex(a, n)); // This code is contributed by ajaykrsharma132.</script> |
Output:
6
Time Complexity: O(N)
Auxiliary Space: O(1)
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