Given an integer K and an array arr[] having N pairwise distinct integers in the range [1, K], the task is to find the lexicographically smallest permutation of the first K positive integers such that the given array arr[] is a subsequence of the permutation.
Examples:
Input: arr[] = {1, 3, 5, 7}, K = 8
Output: 1 2 3 4 5 6 7 8
Explanation: {1, 2, 3, 4, 5, 6, 7, 8} is the lexicographically smallest permutation of the first 8 positive integers such that the given array {1, 3, 5, 7} is also a subsequence of the permutation.Input: arr[] = {6, 4, 2, 1}, K=7
Output: 3 5 6 4 2 1 7
Approach: The given problem can be solved by using a greedy approach. Below are the steps to follow:
- Create a vector missing[] that stores the integers in the range [1, K] in increasing order that are not present in the given array arr[] using the approach discussed in this article.
- Create two pointers p1 and p2 which store the current index in arr[] and missing[] respectively. Initially, both are equal to 0.
- Greedily take and store the minimum of arr[p1] and missing [p2] into a vector, say ans[] and increment the respective pointer to the next position till the count of the stored integers is less than K.
- In order to make things easier, append INT_MAX at the end of the array missing[] and arr[] which will avoid getting out of bounds.
- After completing the above steps, all the values stored in the ans[] is the required result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the lexicographically// smallest permutation such that the// given array is a subsequence of itvoid findPermutation(int K, vector<int> arr){ // Stores the missing elements in // arr in the range [1, K] vector<int> missing; // Stores if the ith element is // present in arr or not vector<bool> visited(K + 1, 0); // Loop to mark all integers present // in the array as visited for (int i = 0; i < arr.size(); i++) { visited[arr[i]] = 1; } // Loop to insert all the integers // not visited into missing for (int i = 1; i <= K; i++) { if (!visited[i]) { missing.push_back(i); } } // Append INT_MAX at end in order // to prevent going out of bounds arr.push_back(INT_MAX); missing.push_back(INT_MAX); // Pointer to the current element int p1 = 0; // Pointer to the missing element int p2 = 0; // Stores the required permutation vector<int> ans; // Loop to construct the permutation // using greedy approach while (ans.size() < K) { // If missing element is smaller // that the current element insert // missing element if (arr[p1] < missing[p2]) { ans.push_back(arr[p1]); p1++; } // Insert current element else { ans.push_back(missing[p2]); p2++; } } // Print the required Permutation for (int i = 0; i < K; i++) { cout << ans[i] << " "; }}// Driver Codeint main(){ int K = 7; vector<int> arr = { 6, 4, 2, 1 }; // Function Call findPermutation(K, arr); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the lexicographically// smallest permutation such that the// given array is a subsequence of itstatic void findPermutation(int K, Vector<Integer> arr){ // Stores the missing elements in // arr in the range [1, K] Vector<Integer> missing = new Vector<Integer>(); // Stores if the ith element is // present in arr or not boolean visited[] = new boolean[K + 1]; // Loop to mark all integers present // in the array as visited for (int i = 0; i < arr.size(); i++) { visited[arr.get(i)] = true; } // Loop to insert all the integers // not visited into missing for (int i = 1; i <= K; i++) { if (!visited[i]) { missing.add(i); } } // Append Integer.MAX_VALUE at end in order // to prevent going out of bounds arr.add(Integer.MAX_VALUE); missing.add(Integer.MAX_VALUE); // Pointer to the current element int p1 = 0; // Pointer to the missing element int p2 = 0; // Stores the required permutation Vector<Integer> ans = new Vector<Integer>(); // Loop to construct the permutation // using greedy approach while (ans.size() < K) { // If missing element is smaller // that the current element insert // missing element if (arr.get(p1) < missing.get(p2)) { ans.add(arr.get(p1)); p1++; } // Insert current element else { ans.add(missing.get(p2)); p2++; } } // Print the required Permutation for (int i = 0; i < K; i++) { System.out.print(ans.get(i)+ " "); }}// Driver Codepublic static void main(String[] args){ int K = 7; Integer []a = {6, 4, 2, 1}; Vector<Integer> arr = new Vector<>(Arrays.asList(a)); // Function Call findPermutation(K, arr);}}// This code is contributed by 29AjayKumar |
Python3
# Python program for the above approach# Function to find the lexicographically# smallest permutation such that the# given array is a subsequence of itdef findPermutation(K, arr): # Stores the missing elements in # arr in the range [1, K] missing = [] # Stores if the ith element is # present in arr or not visited = [0]*(K+1) # Loop to mark all integers present # in the array as visited for i in range(4): visited[arr[i]] = 1 # Loop to insert all the integers # not visited into missing for i in range(1, K+1): if(not visited[i]): missing.append(i) # Append INT_MAX at end in order # to prevent going out of bounds INT_MAX = 2147483647 arr.append(INT_MAX) missing.append(INT_MAX) # Pointer to the current element p1 = 0 # Pointer to the missing element p2 = 0 # Stores the required permutation ans = [] # Loop to construct the permutation # using greedy approach while (len(ans) < K): # If missing element is smaller # that the current element insert # missing element if (arr[p1] < missing[p2]): ans.append(arr[p1]) p1 = p1 + 1 # Insert current element else: ans.append(missing[p2]) p2 = p2 + 1 # Print the required Permutation for i in range(0, K): print(ans[i], end=" ")# Driver Codeif __name__ == "__main__": K = 7 arr = [6, 4, 2, 1] # Function Call findPermutation(K, arr) # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;using System.Collections;class GFG{// Function to find the lexicographically// smallest permutation such that the// given array is a subsequence of itstatic void findPermutation(int K, ArrayList arr){ // Stores the missing elements in // arr in the range [1, K] ArrayList missing = new ArrayList(); // Stores if the ith element is // present in arr or not bool [] visited = new bool[K + 1]; // Loop to mark all integers present // in the array as visited for (int i = 0; i < arr.Count; i++) { visited[(int)arr[i]] = true; } // Loop to insert all the integers // not visited into missing for (int i = 1; i <= K; i++) { if (!visited[i]) { missing.Add(i); } } // Append Int32.MaxValue at end in order // to prevent going out of bounds arr.Add(Int32.MaxValue); missing.Add(Int32.MaxValue); // Pointer to the current element int p1 = 0; // Pointer to the missing element int p2 = 0; // Stores the required permutation ArrayList ans = new ArrayList(); // Loop to construct the permutation // using greedy approach while (ans.Count < K) { // If missing element is smaller // that the current element insert // missing element if ((int)arr[p1] < (int)missing[p2]) { ans.Add(arr[p1]); p1++; } // Insert current element else { ans.Add(missing[p2]); p2++; } } // Print the required Permutation for (int i = 0; i < K; i++) { Console.Write(ans[i]+ " "); }}// Driver Codepublic static void Main(){ int K = 7; int [] a = {6, 4, 2, 1}; ArrayList arr = new ArrayList(a); // Function Call findPermutation(K, arr);}}// This code is contributed by ihritik. |
Javascript
<script> // JavaScript program for the above approach // Function to find the lexicographically // smallest permutation such that the // given array is a subsequence of it const findPermutation = (K, arr) => { // Stores the missing elements in // arr in the range [1, K] let missing = []; // Stores if the ith element is // present in arr or not let visited = new Array(K + 1).fill(0); // Loop to mark all integers present // in the array as visited for (let i = 0; i < arr.length; i++) { visited[arr[i]] = 1; } // Loop to insert all the integers // not visited into missing for (let i = 1; i <= K; i++) { if (!visited[i]) { // missing.push_back(i); missing.push(i); } } // Append INT_MAX at end in order // to prevent going out of bounds const INT_MAX = 2147483647; arr.push(INT_MAX); missing.push(INT_MAX); // Pointer to the current element let p1 = 0; // Pointer to the missing element let p2 = 0; // Stores the required permutation let ans = []; // Loop to construct the permutation // using greedy approach while (ans.length < K) { // If missing element is smaller // that the current element insert // missing element if (arr[p1] < missing[p2]) { ans.push(arr[p1]); p1++; } // Insert current element else { ans.push(missing[p2]); p2++; } } // Print the required Permutation for (let i = 0; i < K; i++) { document.write(`${ans[i]} `); } } // Driver Code let K = 7; let arr = [6, 4, 2, 1]; // Function Call findPermutation(K, arr); // This code is contributed by rakeshsahni.</script> |
Output:
3 5 6 4 2 1 7
Time Complexity: O(N)
Auxiliary Space: O(N)
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