Given an array arr[] of length N such that (1 <= arr[i] <= N), the task is to modify the array, by only inserting elements within the range [1, N], such that the sum of all subarrays of length K become equal. Print the modified array, if possible. Else print “Not possible”.
Examples:
Input: arr[] = {1, 2, 2, 1}, K = 2
Output: 1 2 1 2 1
Explanation:
Insert 1 between second and third element.
Now all subarray of length K has an equal sum of 3.
Input: arr[] = {1, 2, 3, 4}, K = 3
Output: Not possible
Approach:
- Since the removal of elements is not allowed, therefore the only case when such a modified array can be achieved, is when there are at most K distinct elements in the array.
- Therefore, Firstly check if there are more than K distinct elements in the given array. If yes, then print “Not possible”.
- Else, store all the distinct elements of the given array in a vector.
- If the number of distinct elements is less than K, then add/repeat some elements in the vector and make the size of the vector equal to K.
- The vector has now K elements. Now repeat the elements of the vector in the same order to show the modified array. This repetition is done to show all the subarrays of length K have the same sum.
Below is the implementation of the above approach
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function that prints another array// whose all subarray of length k// have an equal sumvoid MakeArray(int a[], int n, int k){ unordered_map<int, int> mp; // Store all distinct elements in // the unordered map for (int i = 0; i < n; i++) { if (mp.find(a[i]) == mp.end()) mp[a[i]] = 1; } // Condition if the number of // distinct elements is greater // than k if (mp.size() > k) { cout << "Not possible\n"; return; } vector<int> ans; // Push all distinct elements // in a vector for (auto i : mp) { ans.push_back(i.first); } // Push 1 while the size of // vector not equal to k while (ans.size() < k) { ans.push_back(1); } // Print the vector 2 times for (int i = 0; i < 2; i++) { for (int j = 0; j < k; j++) cout << ans[j] << " "; }}// Driver Codeint main(){ int arr[] = { 1, 2, 2, 1 }; int K = 2; int size = sizeof(arr) / sizeof(arr[0]); MakeArray(arr, size, K); return 0;} |
Java
// Java implementation of above approachimport java.util.*;class GFG{ // Function that prints another array// whose all subarray of length k// have an equal sumstatic void MakeArray(int a[], int n, int k){ HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Store all distinct elements in // the unordered map for (int i = 0; i < n; i++) { if (!mp.containsKey(a[i])) mp.put(a[i], 1); } // Condition if the number of // distinct elements is greater // than k if (mp.size() > k) { System.out.print("Not possible\n"); return; } Vector<Integer> ans = new Vector<Integer>(); // Push all distinct elements // in a vector for (Map.Entry<Integer,Integer> i : mp.entrySet()) { ans.add(i.getKey()); } // Push 1 while the size of // vector not equal to k while (ans.size() < k) { ans.add(1); } // Print the vector 2 times for (int i = 0; i < 2; i++) { for (int j = 0; j < k; j++) System.out.print(ans.get(j) + " "); }} // Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 2, 1 }; int K = 2; int size = arr.length; MakeArray(arr, size, K);}} // This code is contributed by Rohit_ranjan |
Python3
# Python3 implementation of above approach# Function that prints another array# whose all subarray of length k# have an equal sumdef MakeArray(a, n, k): mp = dict() # Store all distinct elements in # the unordered map for i in a: if i not in mp: mp[a[i]] = 1 # Condition if the number of # distinct elements is greater # than k if(len(mp) > k): print("Not possible") return ans = [] # Push all distinct elements # in a vector for i in mp: ans.append(i) # Push 1 while the size of # vector not equal to k while(len(ans) < k): ans.append(1) # Print the vector 2 times for i in range(2): for j in range(k): print(ans[j], end = " ")# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 2, 1 ] K = 2 size = len(arr) MakeArray(arr, size, K) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of above approachusing System;using System.Collections.Generic;class GFG{// Function that prints another array// whose all subarray of length k// have an equal sumstatic void MakeArray(int []a, int n, int k){ Dictionary<int, int> mp = new Dictionary<int, int>(); // Store all distinct elements in // the unordered map for(int i = 0; i < n; i++) { if (!mp.ContainsKey(a[i])) mp.Add(a[i], 1); } // Condition if the number of // distinct elements is greater // than k if (mp.Count > k) { Console.Write("Not possible\n"); return; } List<int> ans = new List<int>(); // Push all distinct elements // in a vector foreach(KeyValuePair<int, int> i in mp) { ans.Add(i.Key); } // Push 1 while the size of // vector not equal to k while (ans.Count < k) { ans.Add(1); } // Print the vector 2 times for(int i = 0; i < 2; i++) { for(int j = 0; j < k; j++) Console.Write(ans[j] + " "); }}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 2, 1 }; int K = 2; int size = arr.Length; MakeArray(arr, size, K);}}// This code is contributed by amal kumar choubey |
Javascript
<script>// Javascript implementation of above approach// Function that prints another array// whose all subarray of length k// have an equal sumfunction MakeArray(a, n, k){ var mp = new Map(); // Store all distinct elements in // the unordered map for (var i = 0; i < n; i++) { if (!mp.has(a[i])) mp.set(a[i], 1); } // Condition if the number of // distinct elements is greater // than k if (mp.count > k) { document.write( "Not possible<br>"); return; } var ans = []; // Push all distinct elements // in a vector mp.forEach((value, key) => { ans.push(key); }); // Push 1 while the size of // vector not equal to k while (ans.length < k) { ans.push(1); } // Print the vector 2 times for (var i = 0; i < 2; i++) { for (var j = 0; j < k; j++) document.write( ans[j] + " "); }}// Driver Codevar arr = [1, 2, 2, 1];var K = 2;var size = arr.length;MakeArray(arr, size, K);// This code is contributed by rutvik_56.</script> |
Output:
2 1 2 1
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), in order to store elements in HashMap.
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