Check if a key is present in every segment of size k in an array
Given an array arr[] and size of array is n and one another key x, and give you a segment size k. The task is to find that the key x present in every segment of size k in arr[].
Examples:
Input :
arr[] = { 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3}
x = 3
k = 3
Output : Yes
Explanation: There are 4 non-overlapping segments of size k in the array, {3, 5, 2}, {4, 9, 3}, {1, 7, 3} and {11, 12, 3}. 3 is present all segments.Input :
arr[] = { 21, 23, 56, 65, 34, 54, 76, 32, 23, 45, 21, 23, 25}
x = 23
k = 5
Output :Yes
Explanation: There are three segments and last segment is not full {21, 23, 56, 65, 34}, {54, 76, 32, 23, 45} and {21, 23, 25}.
23 is present all window.Input :arr[] = { 5, 8, 7, 12, 14, 3, 9}
x = 8
k = 2
Output : No
Approach:
The idea is simple, we consider every segment of size k and check if x is present in the window or not. We need to carefully handle the last segment.
Below is the implementation of the above approach:
C++
// C++ code to find the every segment size of// array have a search key x#include <bits/stdc++.h>using namespace std;bool findxinkwindowSize(int arr[], int x, int k, int n){ int i; for (i = 0; i < n; i = i + k) { // Search x in segment starting // from index i. int j; for (j = 0; j < k; j++) if (arr[i + j] == x) break; // If loop didn't break if (j == k) return false; } // If n is a multiple of k if (i == n) return true; // Check in last segment if n // is not multiple of k. int j; for (j=i-k; j<n; j++) if (arr[j] == x) break; if (j == n) return false; return true;}// main driverint main(){ int arr[] = { 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 }; int x = 3, k = 3; int n = sizeof(arr) / sizeof(arr[0]); if (findxinkwindowSize(arr, x, k, n)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java code to find the every// segment size of array have// a search key ximport java.util.*;class GFG { static boolean findxinkwindowSize(int N, int[] arr, int x, int k) { int i; boolean b = false; // Iterate from 0 to N - 1 for (i = 0; i < N; i = i + k) { // Iterate from 0 to k - 1 for (int j = 0; j < k; j++) { if (i + j < N && arr[i + j] == x) break; if (j == k) return false; if (i + j >= N) return false; } } if (i >= N) return true; else return b; } // Driver Code public static void main(String args[]) { int arr[] = new int[] { 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 }; int x = 3, k = 3; int n = arr.length; if (findxinkwindowSize(n, arr, x, k)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Vivek258709 |
Python 3
# Python 3 program to find# the every segment size of# array have a search key xdef findxinkwindowSize(arr, x, k, n) : i = 0 while i < n : j = 0 # Search x in segment # starting from index i while j < k : if arr[i + j] == x : break j += 1 # If loop didn't break if j == k : return False i += k # If n is a multiple of k if i == n : return True j = i - k # Check in last segment if n # is not multiple of k. while j < n : if arr[j] == x : break j += 1 if j == n : return False return True# Driver Codeif __name__ == "__main__" : arr = [ 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 ] x, k = 3, 3 n = len(arr) if (findxinkwindowSize(arr, x, k, n)) : print("Yes") else : print("No") # This code is contributed# by ANKITRAI1 |
C#
// C# code to find the every// segment size of array have// a search key xusing System;class GFG{static bool findxinkwindowSize(int[] arr, int x, int k, int n){ int i; for (i = 0; i < n; i = i + k) { // Search x in segment // starting from index i. int j; for (j = 0; j < k; j++) if (arr[i + j] == x) break; // If loop didn't break if (j == k) return false; } // If n is a multiple of k if (i == n) return true; // Check in last segment if // n is not multiple of k. int l; for (l = i - k; l < n; l++) if (arr[l] == x) break; if (l == n) return false; return true;}// Driver Codepublic static void Main(){ int[] arr = new int[] {3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3}; int x = 3, k = 3; int n = arr.Length; if (findxinkwindowSize(arr, x, k, n)) Console.Write("Yes"); else Console.Write("No");}}// This code is contributed by ChitraNayal |
PHP
<?php// PHP code to find the every// segment size of array have// a search key xfunction findxinkwindowSize(&$arr, $x, $k, $n){ for ($i = 0; $i < $n; $i = $i + $k) { // Search x in segment // starting from index i. for ($j = 0; $j < $k; $j++) if ($arr[$i + $j] == $x) break; // If loop didn't break if ($j == $k) return false; } // If n is a multiple of k if ($i == $n) return true; // Check in last segment if n // is not multiple of k. for ($j = $i - $k; $j < $n; $j++) if ($arr[$j] == $x) break; if ($j == $n) return false; return true;}// Driver Code$arr = array(3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3);$x = 3;$k = 3;$n = sizeof($arr);if (findxinkwindowSize($arr, $x, $k, $n)) echo "Yes" ;else echo "No" ;// This code is contributed// by Shivi_Aggarwal?> |
Javascript
<script>// JavaScript code to find the every segment size of// array have a search key xfunction findxinkwindowSize( arr, x, k, n){ let i; for (i = 0; i < n; i = i + k) { // Search x in segment starting // from index i. let j; for (j = 0; j < k; j++) if (arr[i + j] == x) break; // If loop didn't break if (j == k) return false; } // If n is a multiple of k if (i == n) return true; // Check in last segment if n // is not multiple of k. let j; for (j=i-k; j<n; j++) if (arr[j] == x) break; if (j == n) return false; return true;}// main driver let arr = [ 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 ]; let x = 3, k = 3; let n = arr.length; if (findxinkwindowSize(arr, x, k, n)) document.write("Yes"); else document.write("No"); // This code contributed by aashish1995</script> |
Output:
Yes
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1) as constant space is being used.
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