Given two integers N and K and a range [L, R], the task is to build an array whose elements are unique and in the range [L, R] and the median of the array is K.
Examples:
Input: N = 6, K = -1, L = -5, R = 5
Output: -4 -3 -2 0 1 2
Explanation: Median = (-2 + 0)/2 = -1 which is equal to K.Input: N = 5, K = 4, L = 3, R = 8
Output: -1
Approach: The problem can be solved using the following mathematical idea:
- If the array is of odd length the median is the (N/2 + 1)th element after sorting.
- Otherwise, the median will be the average of (N/2) and (N/2 + 1)th elements.
So the optimal way is to make an array such that the minimum value and the maximum value are as close to the median as possible.
This can be done by simply keeping the difference between the elements as 1.
Therefore the minimum element of the array will be (K – N/2) and the maximum element will be (K + N/2).Note: In case of N being even K is the average of two middle elements. So those two elements have a difference of 2 between them and K will not be present in the array.
Follow the steps mentioned below to implement the observation:
- Find the possible minimum and maximum values of the array.
- If either L is greater than the minimum or R is less than the maximum then the array cannot be constructed.
- Otherwise, fix the minimum and generate all the elements by maintaining a gap of 1 between adjacent elements.
- In the case of N being even, the gap between the middle two elements will be 2.
- Return the generated array as the answer.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to construct the arrayvector<int> constructArray(int N, int K, int L, int R){ vector<int> ans; if (L > (K - (N / 2)) || R < (K + (N / 2))) ans.push_back(-1); else if (N & 1) for (int i = K - (N / 2); i <= K + (N / 2); i++) ans.push_back(i); else for (int i = K - (N / 2); i <= K + (N / 2); i++) if (i != K) ans.push_back(i); return ans;}// Driver Codeint main(){ int N = 6, K = -1; int L = -5, R = 5; // Function call vector<int> ans = constructArray(N, K, L, R); for (int x : ans) cout << x << " "; return 0;} |
Java
// Java code to implement the above approachimport java.io.*;import java.util.*;class GFG { // Function to construct the array public static ArrayList<Integer> constructArray(int N, int K, int L, int R) { ArrayList<Integer> ans = new ArrayList<Integer>(); if (L > (K - (N / 2)) || R < (K + (N / 2))) ans.add(-1); else if ((N & 1) != 0) for (int i = K - (N / 2); i <= K + (N / 2); i++) ans.add(i); else for (int i = K - (N / 2); i <= K + (N / 2); i++) if (i != K) ans.add(i); return ans; } // Driver Code public static void main(String[] args) { int N = 6, K = -1; int L = -5, R = 5; // Function call ArrayList<Integer> ans = constructArray(N, K, L, R); for (Integer x : ans) System.out.print(x + " "); }}// This code is contributed by Rohit Pradhan |
Python3
# Python3 code to implement the above approach# Function to construct the arraydef constructArray(N, K, L, R) : ans = []; if (L > (K - (N // 2))) or (R < (K + (N // 2))) : ans.append(-1); elif (N & 1) : for i in range(K- (N//2), K + 1 + (N//2)) : ans.append(i); else : for i in range(K- (N//2), K + 1 + (N//2)) : if (i != K) : ans.append(i); return ans;# Driver Codeif __name__ == "__main__" : N = 6; K = -1; L = -5; R = 5; # Function call ans = constructArray(N, K, L, R); for x in ans: print(x,end= " "); # This code is contributed by AnkThon |
C#
// C# code to implement the above approachusing System;using System.Collections;public class GFG{ // Function to construct the array public static ArrayList constructArray(int N, int K, int L, int R) { ArrayList ans = new ArrayList(); if (L > (K - (N / 2)) || R < (K + (N / 2))) ans.Add(-1); else if ((N & 1) != 0) for (int i = K - (N / 2); i <= K + (N / 2); i++) ans.Add(i); else for (int i = K - (N / 2); i <= K + (N / 2); i++) if (i != K) ans.Add(i); return ans; } // Driver Code static public void Main (){ int N = 6, K = -1; int L = -5, R = 5; // Function call ArrayList ans = constructArray(N, K, L, R); for (int i = 0; i < ans.Count; i++) Console.Write(ans[i] + " "); }}// This code is contributed by Dharanendra L V. |
Javascript
<script>// JavaScript code to implement the above approach// Function to construct the arrayfunction constructArray(N, K, L, R){ let ans = []; if (L > (K - (N / 2)) || R < (K + (N / 2))) ans.push(-1); else if (N & 1) for (let i = K - (N / 2); i <= K + (N / 2); i++) ans.push(i); else{ for (let i = K - (N / 2);i <= K + (N / 2); i++){ if (i != K) ans.push(i); } } return ans;}// Driver Codelet N = 6, K = -1;let L = -5, R = 5;// Function calllet ans = constructArray(N, K, L, R);for(let x of ans) document.write(x + " ");// This code is contributed by shinjanpatra</script> |
Output
-4 -3 -2 0 1 2
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
