# Find an array of size N that satisfies the given conditions

Given three integers N, S and K, the task is to create an array of N positive integers such that the bitwise OR of any two consecutive elements from the array is odd and there are exactly K subarrays with sum equal to S where 1 ≤ K ≤ N / 2.

Examples:

Input: N = 4, K = 2, S = 6
Output: 6 7 6 7
Here, there are exactly 2 subarray {6} and {6}
whose sum is 6 and the bitwise OR of
any adjacent elements is odd.

Input: N = 8, K = 3, S = 12
Output: 12 13 12 13 12 13 13 13

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Observe a pattern here {S, P, S, P, S, P, …, P, P, P, P}.
• Here P is an odd number > S and after every S there is an occurrence of P. It is known that the bitwise OR with an odd number is always an odd number, so it is confirmed that bitwise OR of each adjacent element is an odd number.
• Now, put exactly K number of S in the above patten of the array.
• Except S all the elements (which are P) are greater than S, so there can not be any subarray whose sum is exactly S other than those K subarray.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to print the ` `// contents of an array ` `void` `printArr(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Function to generate and print ` `// the required array ` `void` `findArray(``int` `n, ``int` `k, ``int` `s) ` `{ ` ` `  `    ``// Initially all the positions are empty ` `    ``int` `vis[n] = { 0 }; ` ` `  `    ``// To store the count of positions ` `    ``// i such that arr[i] = s ` `    ``int` `cnt = 0; ` ` `  `    ``// To store the final array elements ` `    ``int` `arr[n]; ` ` `  `    ``for` `(``int` `i = 0; i < n && cnt < k; i += 2) { ` ` `  `        ``// Set arr[i] = s and the gap between ` `        ``// them is exactly 2 so in for loop ` `        ``// we use i += 2 ` `        ``arr[i] = s; ` ` `  `        ``// Mark the i'th position as visited ` `        ``// as we put arr[i] = s ` `        ``vis[i] = 1; ` ` `  `        ``// Increment the count ` `        ``cnt++; ` `    ``} ` `    ``int` `val = s; ` ` `  `    ``// Finding the next odd number after s ` `    ``if` `(s % 2 == 0) ` `        ``val++; ` `    ``else` `        ``val = val + 2; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(vis[i] == 0) { ` ` `  `            ``// If the i'th position is not visited ` `            ``// it means we did not put any value ` `            ``// at position i so we put 1 now ` `            ``arr[i] = val; ` `        ``} ` `    ``} ` ` `  `    ``// Print the final array ` `    ``printArr(arr, n); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 8, k = 3, s = 12; ` ` `  `    ``findArray(n, k, s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `    ``// Utility function to print the  ` `    ``// contents of an array  ` `    ``static` `void` `printArr(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``System.out.print(arr[i] + ``" "``);  ` `    ``}  ` `     `  `    ``// Function to generate and print  ` `    ``// the required array  ` `    ``static` `void` `findArray(``int` `n, ``int` `k, ``int` `s)  ` `    ``{  ` `     `  `        ``// Initially all the positions are empty  ` `        ``int` `vis[] = ``new` `int``[n] ;  ` `     `  `        ``// To store the count of positions  ` `        ``// i such that arr[i] = s  ` `        ``int` `cnt = ``0``;  ` `     `  `        ``// To store the final array elements  ` `        ``int` `arr[] = ``new` `int``[n];  ` `     `  `        ``for` `(``int` `i = ``0``; i < n && cnt < k; i += ``2``)  ` `        ``{  ` `     `  `            ``// Set arr[i] = s and the gap between  ` `            ``// them is exactly 2 so in for loop  ` `            ``// we use i += 2  ` `            ``arr[i] = s;  ` `     `  `            ``// Mark the i'th position as visited  ` `            ``// as we put arr[i] = s  ` `            ``vis[i] = ``1``;  ` `     `  `            ``// Increment the count  ` `            ``cnt++;  ` `        ``}  ` `        ``int` `val = s;  ` `     `  `        ``// Finding the next odd number after s  ` `        ``if` `(s % ``2` `== ``0``)  ` `            ``val++;  ` `        ``else` `            ``val = val + ``2``;  ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `            ``if` `(vis[i] == ``0``) ` `            ``{  ` `     `  `                ``// If the i'th position is not visited  ` `                ``// it means we did not put any value  ` `                ``// at position i so we put 1 now  ` `                ``arr[i] = val;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Print the final array  ` `        ``printArr(arr, n);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `n = ``8``, k = ``3``, s = ``12``;  ` `     `  `        ``findArray(n, k, s);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Utility function to print the  ` `# contents of an array  ` `def` `printArr(arr, n) :  ` ` `  `    ``for` `i ``in` `range``(n) :  ` `        ``print``(arr[i], end``=` `" "``);  ` ` `  `# Function to generate and print  ` `# the required array  ` `def` `findArray(n, k, s) : ` ` `  `    ``# Initially all the positions are empty  ` `    ``vis ``=` `[``0``] ``*` `n;  ` ` `  `    ``# To store the count of positions  ` `    ``# i such that arr[i] = s  ` `    ``cnt ``=` `0``;  ` ` `  `    ``# To store the final array elements  ` `    ``arr ``=` `[``0``] ``*` `n; ` `    ``i ``=` `0``; ` `     `  `    ``while` `(i < n ``and` `cnt < k) : ` `         `  `        ``# Set arr[i] = s and the gap between  ` `        ``# them is exactly 2 so in for loop  ` `        ``# we use i += 2  ` `        ``arr[i] ``=` `s;  ` ` `  `        ``# Mark the i'th position as visited  ` `        ``# as we put arr[i] = s  ` `        ``vis[i] ``=` `1``;  ` ` `  `        ``# Increment the count  ` `        ``cnt ``+``=` `1``; ` `        ``i ``+``=` `2``; ` `    ``val ``=` `s;  ` `     `  `    ``# Finding the next odd number after s  ` `    ``if` `(s ``%` `2` `=``=` `0``) : ` `        ``val ``+``=` `1``;  ` `    ``else` `: ` `        ``val ``=` `val ``+` `2``;  ` ` `  `    ``for` `i ``in` `range``(n) : ` `        ``if` `(vis[i] ``=``=` `0``) : ` ` `  `            ``# If the i'th position is not visited  ` `            ``# it means we did not put any value  ` `            ``# at position i so we put 1 now  ` `            ``arr[i] ``=` `val;  ` ` `  `    ``# Print the final array  ` `    ``printArr(arr, n);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``n ``=` `8``; k ``=` `3``; s ``=` `12``;  ` ` `  `    ``findArray(n, k, s);  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Utility function to print the  ` `    ``// contents of an array  ` `    ``static` `void` `printArr(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``Console.Write(arr[i] + ``" "``);  ` `    ``}  ` `     `  `    ``// Function to generate and print  ` `    ``// the required array  ` `    ``static` `void` `findArray(``int` `n, ``int` `k, ``int` `s)  ` `    ``{  ` `     `  `        ``// Initially all the positions are empty  ` `        ``int` `[]vis = ``new` `int``[n] ;  ` `     `  `        ``// To store the count of positions  ` `        ``// i such that arr[i] = s  ` `        ``int` `cnt = 0;  ` `     `  `        ``// To store the final array elements  ` `        ``int` `[]arr = ``new` `int``[n];  ` `     `  `        ``for` `(``int` `i = 0; i < n && cnt < k; i += 2)  ` `        ``{  ` `     `  `            ``// Set arr[i] = s and the gap between  ` `            ``// them is exactly 2 so in for loop  ` `            ``// we use i += 2  ` `            ``arr[i] = s;  ` `     `  `            ``// Mark the i'th position as visited  ` `            ``// as we put arr[i] = s  ` `            ``vis[i] = 1;  ` `     `  `            ``// Increment the count  ` `            ``cnt++;  ` `        ``}  ` `        ``int` `val = s;  ` `     `  `        ``// Finding the next odd number after s  ` `        ``if` `(s % 2 == 0)  ` `            ``val++;  ` `        ``else` `            ``val = val + 2;  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `            ``if` `(vis[i] == 0)  ` `            ``{  ` `     `  `                ``// If the i'th position is not visited  ` `                ``// it means we did not put any value  ` `                ``// at position i so we put 1 now  ` `                ``arr[i] = val;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Print the final array  ` `        ``printArr(arr, n);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `n = 8, k = 3, s = 12;  ` `     `  `        ``findArray(n, k, s);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01  `

Output:

```12 13 12 13 12 13 13 13
```

Time Complexity: O(N)

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Improved By : AnkitRai01