Given N people standing in a queue and two arrays A[] and B[]. The array A[] represent the name of the person and array B[] represents how many people are taller than a particular person standing in front of that. Now the queue is shuffled. The task is to print the original sequence of the queue following the above property.
Examples:
Input: N = 4, A[] = {‘a’, ‘b’, ‘c’, ‘d’}, B[] = {0, 2, 0, 0}
Output:
a 1
c 3
d 4
b 2
Explanation:
Looking at the output queue and their generated heights, it can be easily understood that:
1) a is the first one in the queue and so we have the person with 0th index in front of him. So a is associated with 0 in the input.
2) c has only a in front of him/her but a is shorter than c. Therefore c is associated with 0 in the input.
3) d has c and a in front of him/her but they are both shorter than d . Therefore d is associated with 0 in the input.
4) b has d, c and a in front of b. But only c and d are taller than b. So, b is associated with 2 in the input.Input: N = 4, A[] = { ‘a’, ‘b’, ‘c’, ‘d’}, B[] = { 0, 1, 3, 3}
Output: -1
Explanation:
The given order is the original order of the queue.
Approach:
- Firstly make a pair of the person’s name and their associated integers and sort the pairs.
- Create an array answer[] to store the possible heights of the person.
- Iterate over all the pair and if the number of persons standing in front of is taller and is greater than their current standing position, then return -1.
- Otherwise, store the difference between the current standing position and the height of the person taller than him, in the answer array.
- For every person iterate over the pair and if the value of the answer array for our current person is greater than the person with whom we are comparing, increment in the answer array for current pair.
- Finally, print the possible pairs from the given sequence according to values stored in answer[] array.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to generate the Queuevoid OriginalQueue(char A[], int B[], int N){ // Making a pair pair<int, string> a[N + 1]; // Answer array int ans[N + 1]; bool possible = true; // Store the values in the pair for (int i = 0; i < N; i++) { a[i].second = A[i]; a[i].first = B[i]; } // Sort the pair sort(a, a + N); // Traverse the pairs for (int i = 0; i < N; i++) { int len = i - a[i].first; // If it is not possible to // generate the Queue if (len < 0) { cout << "-1"; possible = false; } if (!possible) break; else { ans[i] = len; for (int j = 0; j < i; j++) { // Increment the answer if (ans[j] >= ans[i]) ans[j]++; } } // Finally printing the answer if (i == N - 1 && possible) { for (int i = 0; i < N; i++) { cout << a[i].second << " " << ans[i] + 1 << endl; } } }}// Driver Codeint main(){ int N = 4; // Given name of person as char char A[N] = { 'a', 'b', 'c', 'd' }; // Associated integers int B[N] = { 0, 2, 0, 0 }; // Function Call OriginalQueue(A, B, N); return 0;} |
Java
// Java program for the above approach import java.util.*;import java.io.*;class GFG{ // Function to generate the Queuestatic void OriginalQueue(char A[], int B[], int N){ // Making a pair int[][] a = new int[N][2]; // Answer array int[] ans = new int[N]; boolean possible = true; // Store the values in the pair for(int i = 0; i < N; i++) { a[i][0] = B[i]; a[i][1] = (int)A[i]; } // Sort the pair Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]); // Traverse the pairs for(int i = 0; i < N; i++) { int len = i - a[i][0]; // If it is not possible to // generate the Queue if (len < 0) { System.out.print("-1"); possible = false; } if (!possible) break; else { ans[i] = len; for(int j = 0; j < i; j++) { // Increment the answer if (ans[j] >= ans[i]) ans[j]++; } } // Finally printing the answer if (i == N - 1 && possible) { for(int k = 0; k < N; k++) { System.out.println((char)a[k][1] + " "+ (ans[k] + 1)); } } }}// Driver Codepublic static void main (String[] args){ int N = 4; // Given name of person as char char A[] = { 'a', 'b', 'c', 'd' }; // Associated integers int B[] = { 0, 2, 0, 0 }; // Function Call OriginalQueue(A, B, N);}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Function to generate the Queuedef OriginalQueue(A, B, N): # Making a pair a = [[0, ""] for i in range(N)] # Answer array ans = [0 for i in range(N)] possible = True # Store the values in the pair for i in range(N): a[i][1] = str(A[i]) a[i][0] = B[i] # Sort the pair a.sort(reverse = False) # Traverse the pairs for i in range(N): len1 = i - a[i][0] # If it is not possible to # generate the Queue if (len1 < 0): print("-1",end = "") possible = False if (possible == False): break else: ans[i] = len1 for j in range(i): # Increment the answer if (ans[j] >= ans[i]): ans[j] += 1 # Finally printing the answer if (i == N - 1 and possible): for i in range(N): print(a[i][1], ans[i] + 1)# Driver Codeif __name__ == '__main__': N = 4 # Given name of person as char A = [ 'a', 'b', 'c', 'd' ] # Associated integers B = [ 0, 2, 0, 0 ] # Function Call OriginalQueue(A, B, N)# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;using System.Linq;public class GFG { // Function to generate the Queue static void OriginalQueue(char[] A, int[] B, int N) { // Making a pair int[][] a = new int[N][]; for(int i = 0; i < N; i++) { a[i] = new int[] {B[i], (int)A[i]}; } // Answer array int[] ans = new int[N]; bool possible = true; // Sort the pair Array.Sort(a, (o1, o2) => o1[0] - o2[0]); // Traverse the pairs for(int i = 0; i < N; i++) { int len = i - a[i][0]; // If it is not possible to generate the Queue if (len < 0) { Console.Write("-1"); possible = false; } if (!possible) break; else { ans[i] = len; for(int j = 0; j < i; j++) { // Increment the answer if (ans[j] >= ans[i]) ans[j]++; } } // Finally printing the answer if (i == N - 1 && possible) { for(int k = 0; k < N; k++) { Console.WriteLine((char)a[k][1] + " " + (ans[k] + 1)); } } } } // Driver Code public static void Main(string[] args) { int N = 4; // Given name of person as char char[] A = {'a', 'b', 'c', 'd'}; // Associated integers int[] B = {0, 2, 0, 0}; // Function Call OriginalQueue(A, B, N); }}// This code is contributed by princekumaras |
Javascript
<script>// Javascript program for the above approach// Function to generate the Queuefunction OriginalQueue(A, B, N){ // Making a pair var a = Array(N + 1); for(var i = 0; i < N; i++) { a[i] = [0, ""]; } // Answer array var ans = Array(N + 1); var possible = true; // Store the values in the pair for(var i = 0; i < N; i++) { a[i][1] = A[i]; a[i][0] = B[i]; } // Sort the pair a.sort() // Traverse the pairs for(var i = 0; i < N; i++) { var len = i - a[i][0]; // If it is not possible to // generate the Queue if (len < 0) { document.write("-1"); possible = false; } if (!possible) break; else { ans[i] = len; for(var j = 0; j < i; j++) { // Increment the answer if (ans[j] >= ans[i]) ans[j]++; } } // Finally printing the answer if (i == N - 1 && possible) { for(var i = 0; i < N; i++) { document.write(a[i][1] + " " + (ans[i] + 1) + "<br>"); } } }}// Driver Codevar N = 4;// Given name of person as charvar A = [ 'a', 'b', 'c', 'd' ];// Associated integersvar B = [ 0, 2, 0, 0 ];// Function CallOriginalQueue(A, B, N);// This code is contributed by itsok</script> |
Output:
a 1 c 3 d 4 b 2
Time Complexity: O(N2)
Auxiliary Space: O(N)
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