Given integers N and K, the task is to construct an array arr[] of size N using numbers in the range [1, N] such that it has K sub-arrays whose all the elements are distinct.
Note: If there are multiple possible answers return any of them.
Examples:
Input: N = 5, K = 8
Output: {1, 2, 3, 3, 3}
Explanation: This array has the distinct sub-arrays as {1}, {2}, {3}, {3}, {3}, {1, 2}, {2, 3}, {1, 2, 3}Input: N = 6, K = 21
Output: {1, 2, 3, 4, 5, 6}
Explanation: This array has the 21 distinct sub-arrays.
Approach: The idea to solve the problem is based on the following mathematical observation:
- N elements will definitely form N subarrays of size 1 having unique elements.
- Now the remaining task is to form array such that (N-K subarrays) of size more than 1 have all distinct elements.
- Also it is known number of subarrays of size more than 1 from X elements are (X*(X+1))/2 – X = X*(X-1)/2.
- If X elements are distinct all these subarrays have all distinct elements.
So to form the array there is a need for such X distinct elements such that X*(X-1)/2 = K-N.
So in each step, add a distinct element until the above condition is satisfied. After that repeat the last element till the array size becomes N (because if the last element is repeated it will not effect count of subarrays with all distinct elements).
Follow these steps to solve the problem:
- Decrement K by K – N because every integer contributes to a distinct subarray of size 1.
- Now Initialize a variable num = 0 to keep track of integers that are being added to the array and the number of subarrays formed.
- From K, decrement the number of distinct subarrays formed after adding (num + 1) to the new array. (till num <= K).
- Check if array size has reached N. If not add the num – K repeated times till the array fills.
Below is the implementation for the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to construct the array// with k distinct subarraysvector<int> construct_array(int n, int k){ // Each individual integers // contributes to one subarray k = k - n; // Initialize a variable to keep // track of integers that are // adding to the array int num = 0; // Initialize the res to // store the array vector<int> res; // Push as many possible distinct // integers into the vector while (k >= num) { res.push_back(num + 1); // Decrement the count of // distinct subarrays incrementing k -= num; num++; } // If still it is not possible to // get the array of size n then // adding n-k at the end make num-k // more distinct subarrays while (res.size() < n) { res.push_back(num - k); } // Return the array return res;}// Driver codeint main(){ int N = 5; int K = 8; // Function call vector<int> ans = construct_array(N, K); for (int x : ans) cout << x << " "; return 0;} |
Java
// JAVA program for the above approachimport java.util.*;class GFG { // Function to construct the array // with k distinct subarrays public static ArrayList<Integer> construct_array(int n, int k) { // Each individual integers // contributes to one subarray k = k - n; // Initialize a variable to keep // track of integers that are // adding to the array int num = 0; // Initialize the res to // store the array ArrayList<Integer> res = new ArrayList<Integer>(); // Push as many possible distinct // integers into the vector while (k >= num) { res.add(num + 1); // Decrement the count of // distinct subarrays incrementing k -= num; num++; } // If still it is not possible to // get the array of size n then // adding n-k at the end make num-k // more distinct subarrays while (res.size() < n) { res.add(num - k); } // Return the array return res; } // Driver code public static void main(String[] args) { int N = 5; int K = 8; // Function call ArrayList<Integer> ans = construct_array(N, K); for (int x = 0; x < ans.size(); x++) System.out.print(ans.get(x) + " "); }}// This code is contributed by Taranpreet |
Python3
# Python3 code to implement the approach# Function to construct the array# with k distinct subarraysdef construct_array(n, k): # Each individual integers # contributes to one subarray k -= n # // Initialize a variable to keep # track of integers that are # adding to the array num = 0 # Initialize the res to # store the array res = [] # Push as many possible distinct # integers into the vector while k >= num: res.append(num + 1) # Decrement the count of # distinct subarrays incrementing k -= num num += 1 # If still it is not possible to # get the array of size n then # adding n-k at the end make num-k # more distinct subarrays while len(res) < n: res.append(num - k) return res# Driver codeN = 5K = 8# function callans = construct_array(N, K)print(" ".join(list(map(str, ans))))# This code is contributed by phasing17 |
C#
// C# program for the above approachusing System;using System.Collections;public class GFG{ // Function to construct the array // with k distinct subarrays public static ArrayList construct_array(int n, int k) { // Each individual integers // contributes to one subarray k = k - n; // Initialize a variable to keep // track of integers that are // adding to the array int num = 0; // Initialize the res to // store the array ArrayList res = new ArrayList(); // Push as many possible distinct // integers into the vector while (k >= num) { res.Add(num + 1); // Decrement the count of // distinct subarrays incrementing k -= num; num++; } // If still it is not possible to // get the array of size n then // adding n-k at the end make num-k // more distinct subarrays while (res.Count < n) { res.Add(num - k); } // Return the array return res; } // Driver code static public void Main (){ int N = 5; int K = 8; // Function call ArrayList ans = construct_array(N, K); foreach(int x in ans) Console.Write(x + " "); }}// This code is contributed by hrithikgarg03188. |
Javascript
<script> // JavaScript program for the above approach // Function to construct the array // with k distinct subarrays const construct_array = (n, k) => { // Each individual integers // contributes to one subarray k = k - n; // Initialize a variable to keep // track of integers that are // adding to the array let num = 0; // Initialize the res to // store the array let res = []; // Push as many possible distinct // integers into the vector while (k >= num) { res.push(num + 1); // Decrement the count of // distinct subarrays incrementing k -= num; num++; } // If still it is not possible to // get the array of size n then // adding n-k at the end make num-k // more distinct subarrays while (res.length < n) { res.push(num - k); } // Return the array return res; } // Driver code let N = 5; let K = 8; // Function call let ans = construct_array(N, K); for (let x in ans) document.write(`${ans[x]} `);// This code is contributed by rakeshsahni</script> |
Output
1 2 3 3 3
Time Complexity: O(N)
Auxiliary Space: O(N)
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