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.

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++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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>

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.