Find final value if we double after every successful search in array
Last Updated : 19 Sep, 2023
Given an array and an integer k, traverse the array and if the element in array is k, double the value of k and continue traversal. In the end return value of k.
Examples:
Input : arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2 Output: 16 Explanation: First k = 2 is found, then we search for 4 which is also found, then we search for 8 which is also found, then we search for 16. Input : arr[] = { 2, 4, 5, 6, 7 }, k = 3 Output: 3 Method - 1: (Brute-force)
- Traverse each element of an array if arr[i] == k then k = 2 * k.
- Repeat the same process for the max value of k.
- At last Return the value of k.
Implementation:
C++ // C++ program to find value if we double // the value after every successful search #include <bits/stdc++.h> using namespace std; // Function to Find the value of k int findValue(int a[], int n, int k) { bool exist = true; // Search for k. After every successful // search, double k and change exist to true // and search again for k from the start of array while(exist){ exist = false; for (int i = 0; i < n; i++) { // Check is a[i] is equal to k if (a[i] == k){ k *= 2; exist = true; break; } } } return k; } // Driver's Code int main() { int arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2; int n = sizeof(arr) / sizeof(arr[0]); cout << findValue(arr, n, k); return 0; } // Java program to find value // if we double the value after // every successful search class GFG { // Function to Find the value of k static int findValue(int arr[], int n, int k) { boolean exist = true; // Search for k. After every successful // search, double k and change exist to true // and search again for k from the start of array while(exist){ exist = false; for (int i = 0; i < n; i++) { // Check is a[i] is equal to k if (arr[i] == k){ k *= 2; exist = true; break; } } } return k; } // Driver Code public static void main(String[] args) { int arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2; int n = arr.length; System.out.print(findValue(arr, n, k)); } } // This code is contributed by Aarti_Rathi using System; /* C# program to find value if we double the value after every successful search */ public class GFG { // Function to Find the value of k static int findValue(int[] a, int n, int k) { bool exist = true; // Search for k. After every successful // search, double k and change exist to true // and search again for k from the start of array while (exist) { exist = false; for (int i = 0; i < n; i++) { // Check is a[i] is equal to k if (a[i] == k) { k *= 2; exist = true; break; } } } return k; } // Driver Code public static void Main() { int[] arr = { 2, 3, 4, 10, 8, 1 }; int k = 2; int n = arr.Length; Console.WriteLine(findValue(arr, n, k)); } } // This code is contributed by Aarti_Rathi # Python program to find value if we double # the value after every successful search # Function to Find the value of k def findValue(a, n, k): exist = True while exist: # Search for k. After every successful # search, double k and change exist to true # and search again for k from the start of array exist = False for i in range(n): # Check is a[i] is equal to k if a[i] == k: k *= 2 exist = True break return k # Driver's Code arr = [2, 3, 4, 10, 8, 1] k = 2 n = len(arr) print(findValue(arr, n, k))
<script> // JavaScript program to find value // if we double the value after // every successful search // Function to Find the value of k function findValue(arr, n, k) { var exist = true; // Search for k. After every successful // search, double k and change exist to true // and search again for k from the start of array while(exist){ exist = false; for(let i = 0; i < n; i++){ // Check is a[i] is equal to k if(arr[i] == k) { k *= 2; exist = true; break; } } } return k; } // Driver code let arr = [ 2, 3, 4, 10, 8, 1 ], k = 2; let n = arr.length; document.write(findValue(arr, n, k)); // This code is contributed by muditj148. </script> Time Complexity : O(n^2)
Auxiliary Space: O(1)
Method - 2: (Sort and the search)
- Sort the array
- Then you can just search for the element in one loop because we are sure that k*2 would be after k in this array. Therefore, just multiply the value of k there in the loop only.
Implementation:
C++ // CPP program to find value if we double // the value after every successful search #include <bits/stdc++.h> using namespace std; // Function to Find the value of k int findValue(int a[], int n, int k) { // Sort the array sort(a, a + n); // Search for k. After every successful // search, double k. for (int i = 0; i < n; i++) { // Check is a[i] is equal to k if (a[i] == k) k *= 2; } return k; } // Driver's Code int main() { int arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2; int n = sizeof(arr) / sizeof(arr[0]); cout << findValue(arr, n, k); return 0; } // Java program to find value // if we double the value after // every successful search class GFG { // Function to Find the value of k static int findValue(int arr[], int n, int k) { // Search for k. After every successful // search, double k. for (int i = 0; i < n; i++) if (arr[i] == k) k *= 2; return k; } // Driver Code public static void main(String[] args) { int arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2; int n = arr.length; System.out.print(findValue(arr, n, k)); } } // This code is contributed by // Smitha Dinesh Semwal # Python program to find # value if we double # the value after every # successful search # Function to Find the value of k def findValue(arr, n, k): # Search for k. # After every successful # search, double k. for i in range(n): if (arr[i] == k): k = k * 2 return k # Driver's Code arr = [2, 3, 4, 10, 8, 1] k = 2 n = len(arr) print(findValue(arr, n, k)) # This code is contributed # by Anant Agarwal.
// C# program to find value // if we double the value after // every successful search using System; class GFG { // Function to Find the value of k static int findValue(int[] arr, int n, int k) { // Search for k. After every successful // search, double k. for (int i = 0; i < n; i++) if (arr[i] == k) k *= 2; return k; } // Driver Code public static void Main() { int[] arr = { 2, 3, 4, 10, 8, 1 }; int k = 2; int n = arr.Length; Console.WriteLine(findValue(arr, n, k)); } } // This code is contributed by vt_m. <?php // PHP program to find // value if we double // the value after every // successful search // Function to Find // the value of k function findValue($arr, $n, $k) { // Search for k. After every // successful search, double k. for ($i = 0; $i < $n; $i++) if ($arr[$i] == $k) $k *= 2; return $k; } // Driver Code $arr = array(2, 3, 4, 10, 8, 1); $k = 2; $n = count($arr); echo findValue($arr, $n, $k); // This code is contributed by anuj_67. ?> <script> // JavaScript program to find value // if we double the value after // every successful search // Function to Find the value of k function findValue(arr, n, k) { // Search for k. After every successful // search, double k. for (let i = 0; i < n; i++) if (arr[i] == k) k *= 2; return k; } // Driver code let arr = [ 2, 3, 4, 10, 8, 1 ], k = 2; let n = arr.length; document.write(findValue(arr, n, k)); </script> Time Complexity : O(nlogn)
Auxiliary Space: O(1)
Method - 3: (Hashing)
- Put all elements in hashmap.
- Search if k is in hashmap, If it is then multiply the value by k or return value of k.
Implementation:
C++ // CPP program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the value int findValue(int a[], int n, int k) { // Unordered Map unordered_set<int> m; // Iterate from 0 to n - 1 for (int i = 0; i < n; i++) m.insert(a[i]); while (m.find(k) != m.end()) k = k * 2; return k; } // Driver's Code int main() { int arr[] = { 2, 3, 4, 10, 8, 1 }, k = 2; int n = sizeof(arr) / sizeof(arr[0]); cout << findValue(arr, n, k); return 0; } /*package whatever //do not write package name here */ import java.io.*; import java.util.*; class GFG { static int findValue(int[] a,int n,int k){ // Unordered set HashSet<Integer> m = new HashSet<Integer>(); // Iterate from 0 to n - 1 for(int i=0;i<n;i++){ m.add(a[i]); } while (m.contains(k)){ k = k * 2; } return k; } // Drivers code public static void main(String args[]){ int[] arr = { 2, 3, 4, 10, 8, 1 }; int k = 2; int n = arr.length; System.out.println(findValue(arr, n, k)); } } // This code is contributed by shinjanpatra. # Python program for the above approach # Function to find the value def findValue(a, n, k): # Unordered Map m = set() # Iterate from 0 to n - 1 for i in range(n): m.add(a[i]) while (k in m): k = k * 2 return k # Driver's Code arr, k = [ 2, 3, 4, 10, 8, 1 ], 2 n = len(arr) print(findValue(arr, n, k)) # This code is contributed by shinjanpatra
// C# program for the above approach using System; using System.Collections.Generic; class GFG { static int findValue(int[] a, int n, int k) { // Unordered set HashSet<int> m = new HashSet<int>(); // Iterate from 0 to n - 1 for (int i = 0; i < n; i++) { m.Add(a[i]); } while (m.Contains(k)) { k = k * 2; } return k; } // Drivers code public static void Main(string[] args) { int[] arr = { 2, 3, 4, 10, 8, 1 }; int k = 2; int n = arr.Length; Console.WriteLine(findValue(arr, n, k)); } } // This code is contributed by Tapesh(tapeshdua420) <script> // JavaScript program for the above approach // Function to find the value function findValue(a, n, k) { // Unordered Map let m = new Set(); // Iterate from 0 to n - 1 for (let i = 0; i < n; i++) m.add(a[i]); while (m.has(k)) k = k * 2; return k; } // Driver's Code let arr = [ 2, 3, 4, 10, 8, 1 ], k = 2; let n = arr.length; document.write(findValue(arr, n, k)); // This code is contributed by shinjanpatra </script> Time Complexity: O(n)
Space Complexity: O(n)
Reference: "https://www.geeksforgeeks.org/flipkart-interview-experience-set-35-on-campus-for-sde-1/"
Similar Reads
Find if array has an element whose value is half of array sum Given a sorted array (with unique entries), we have to find whether there exists an element(say X) that is exactly half the sum of all the elements of the array including X. Examples: Input : A = {1, 2, 3} Output : YES Sum of all the elements is 6 = 3*2; Input : A = {2, 4} Output : NO Sum of all the
10 min read
Find a String in given Array of Strings using Binary Search Given a sorted array of Strings arr and a string x, The task is to find the index of x in the array using the Binary Search algorithm. If x is not present, return -1.Examples:Input: arr[] = {"contribute", "geeks", "ide", "practice"}, x = "ide"Output: 2Explanation: The String x is present at index 2.
6 min read
Find elements of original array from doubled array Given an array arr[] of 2*N integers such that it consists of all elements along with the twice values of another array, say A[], the task is to find the array A[]. Examples: Input: arr[] = {4, 1, 18, 2, 9, 8}Output: 1 4 9Explanation:After taking double values of 1, 4, and 9 then adding them to the
7 min read
Find a Fixed Point (Value equal to index) in a given array | Duplicates Allowed Given an array of n integers sorted in ascending order, write a function that returns a Fixed Point in the array, if there is any Fixed Point present in the array, else returns -1. Fixed Point in an array is an index i such that arr[i] is equal to i. Note that integers in the array can be negative.
13 min read
Count of elements that are binary searchable in the given array Given an array arr[] consisting of N integers, the task is to find the maximum count of integers that are binary searchable in the given array. Examples: Input: arr[] = {1, 3, 2}Output: 2Explanation: arr[0], arr[1] can be found. Input: arr[] = {3, 2, 1, 10, 23, 22, 21}Output: 3Explanation: arr[1], a
13 min read