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
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
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
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 findxinkindowSize(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 driver int 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 (findxinkindowSize(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 x class GFG { static boolean findxinkindowSize(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; } // 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 (findxinkindowSize(arr, x, k, n)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by Kirti_Mangal |
Python 3
# Python 3 program to find # the every segment size of # array have a search key x def findxinkindowSize(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 Code if __name__ == "__main__" : arr = [ 3, 5, 2, 4, 9, 3, 1, 7, 3, 11, 12, 3 ] x, k = 3, 3 n = len(arr) if (findxinkindowSize(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 x using System; class GFG { static bool findxinkindowSize(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 Code public 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 (findxinkindowSize(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 x function findxinkindowSize(&$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 (findxinkindowSize($arr, $x, $k, $n)) echo "Yes" ; else echo "No" ; // This code is contributed // by Shivi_Aggarwal ?> |
Yes
Time Complexity: O(n)
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