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:
Java
// 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 import java.util.*; class GFG { static boolean findxinkindowSize( 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 (findxinkindowSize(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 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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