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.
Algorithm:
Step 1: Create a function named “findxinkwindowSize” which takes input parameters “N “- the size of the array, “arr” – the input array, “x” – the search key, “k” – the segment size.
Step 2: Create a boolean variable and initialize it to false
Step 3: Traverse i from 0 to N-1 in steps of k.
Step 4: Now, for each traversed i , iterate j from 0 to k-1 and perform the following:
a. Check if the (i+j)th element of the array arr is equal to x. If yes, break out of the inner loop.
b. If j equals k, return false.
c. If (i+j) is greater than or equal to N, return false.
Step 5: Return true if I is more than or equal to N; else, return b’s value.
Below is the implementation of the above approach:
// 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 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 (findxinkwindowSize(arr, x, k, n))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} |
// Java code to find the every // segment size of array have // a search key x import 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 program to find # the every segment size of # array have a search key x def 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 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 (findxinkwindowSize(arr, x, k, n)) :
print ( "Yes" )
else :
print ( "No" )
# This code is contributed # by ANKITRAI1 |
// C# code to find the every // segment size of array have // a search key x using 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 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 (findxinkwindowSize(arr, x, k, n))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by ChitraNayal |
<?php // PHP code to find the every // segment size of array have // a search key x function 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 ?> |
<script> // JavaScript code to find the every segment size of // array have a search key x function 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> |
Yes
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1) as constant space is being used.