Check if a key is present in every segment of size k in an array
Last Updated :
15 Mar, 2023
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++
#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) {
int j;
for (j = 0; j < k; j++)
if (arr[i + j] == x)
break;
if (j == k)
return false;
}
if (i == n)
return true;
int j;
for (j=i-k; j<n; j++)
if (arr[j] == x)
break;
if (j == n)
return false;
return true;
}
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
import java.util.*;
class GFG {
static boolean findxinkwindowSize(int N, int[] arr,
int x, int k)
{
int i;
boolean b = false;
for (i = 0; i < N; i = i + k) {
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;
}
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");
}
}
|
Python 3
def findxinkwindowSize(arr, x, k, n) :
i = 0
while i < n :
j = 0
while j < k :
if arr[i + j] == x :
break
j += 1
if j == k :
return False
i += k
if i == n :
return True
j = i - k
while j < n :
if arr[j] == x :
break
j += 1
if j == n :
return False
return True
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")
|
C#
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)
{
int j;
for (j = 0; j < k; j++)
if (arr[i + j] == x)
break;
if (j == k)
return false;
}
if (i == n)
return true;
int l;
for (l = i - k; l < n; l++)
if (arr[l] == x)
break;
if (l == n)
return false;
return true;
}
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");
}
}
|
PHP
<?php
function findxinkwindowSize(&$arr, $x,
$k, $n)
{
for ($i = 0;
$i < $n; $i = $i + $k)
{
for ($j = 0; $j < $k; $j++)
if ($arr[$i + $j] == $x)
break;
if ($j == $k)
return false;
}
if ($i == $n)
return true;
for ($j = $i - $k; $j < $n; $j++)
if ($arr[$j] == $x)
break;
if ($j == $n)
return false;
return true;
}
$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" ;
?>
|
Javascript
<script>
function findxinkwindowSize( arr, x, k, n)
{
let i;
for (i = 0; i < n; i = i + k) {
let j;
for (j = 0; j < k; j++)
if (arr[i + j] == x)
break;
if (j == k)
return false;
}
if (i == n)
return true;
let j;
for (j=i-k; j<n; j++)
if (arr[j] == x)
break;
if (j == n)
return false;
return true;
}
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");
</script>
|
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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