# Generate an N-length array having sum of each subarray divisible by K

Given two positive integers** N** and **K**, the task is to generate an array consisting of **N** distinct integers such that the sum of elements of each subarray of the constructed array is divisible by **K**.

**Examples:**

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Input:N = 3, K = 3Output:3 6 9Explanation:

The subarrays of the resultant array are {3}, {6}, {3, 6}, {9}, {6, 9}, {3, 6, 9}. Sum of all these subarrayx are divisible by K.

Input:N = 5, K = 1Output:1 2 3 4 5

**Approach: **Follow the steps below to solve the problem:

- Since the sum of elements of each subarray needs to be divisible by
**K**, the most optimal approach would be to construct an array with each element being a multiple of**K**. - Therefore, iterate a loop from
**i = 1**to**i = N**and for each value of**i**, print**K * i**.

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to construct an array` `// with sum of each subarray` `// divisible by K` `void` `construct_Array(` `int` `N, ` `int` `K)` `{` ` ` `// Traverse a loop from 1 to N` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` `// Print i-th multiple of K` ` ` `cout << K * i << ` `" "` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 3, K = 3;` ` ` `construct_Array(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG` `{` ` ` `// Function to construct an array` ` ` `// with sum of each subarray` ` ` `// divisible by K` ` ` `static` `void` `construct_Array(` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Traverse a loop from 1 to N` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++)` ` ` `{` ` ` `// Print i-th multiple of K` ` ` `System.out.print(K * i + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `3` `, K = ` `3` `;` ` ` `construct_Array(N, K);` ` ` `}` `}` `// This code is contributed by code hunt.` |

## Python3

`# Python program for the above approach` `# Function to construct an array` `# with sum of each subarray` `# divisible by K` `def` `construct_Array(N, K) :` ` ` ` ` `# Traverse a loop from 1 to N` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `+` `1` `):` ` ` `# Pri-th multiple of K` ` ` `print` `(K ` `*` `i, end ` `=` `" "` `)` ` ` `# Driver Code` `N ` `=` `3` `K ` `=` `3` `construct_Array(N, K)` `# This code is contributed by splevel62.` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `public` `class` `GFG` `{` ` ` `// Function to construct an array` ` ` `// with sum of each subarray` ` ` `// divisible by K` ` ` `static` `void` `construct_Array(` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Traverse a loop from 1 to N` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `{` ` ` `// Print i-th multiple of K` ` ` `Console.Write(K * i + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `int` `N = 3, K = 3;` ` ` `construct_Array(N, K);` ` ` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to construct an array` `// with sum of each subarray` `// divisible by K` `function` `construct_Array(N, K)` `{` ` ` `// Traverse a loop from 1 to N` ` ` `for` `(let i = 1; i <= N; i++) {` ` ` `// Print i-th multiple of K` ` ` `document.write( K * i + ` `" "` `);` ` ` `}` `}` `// Driver Code` ` ` `let N = 3, K = 3;` ` ` `construct_Array(N, K);` `// This code is contributed by todaysgaurav` `</script>` |

**Output:**

3 6 9

**Time Complexity:** O(N)**Auxiliary Space:** O(1)