Find the distance between two person after reconstruction of queue

Given a 2D array of size N containing distinct height of people standing in a queue, and a number corresponding to each person (P) that gives the number of persons who are shorter than P and standing in front of P. The task is to determine the distance between two people A and B in the actual order of people’s height.

Example:

Input: A = 6, B = 5, arr[][] = {{5, 0}, {3, 0}, {2, 0},
{6, 4}, {1, 0}, {4, 3}}
Output: 4
Actual arrangement of people’s height
{5, 0}, {3, 0}, {2, 0}, {1, 0}, {6, 4}, {4, 3}
Distance between 6 and 5 is 4

Input: A = 1, B = 3, arr[][] = {{3, 0}, {1, 0}, {2, 1}};
Output: 1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: A brute force approach is to try out all possible permutation of the given heights of the order of the phoenix members, and verify if the in front numbers match for the given sequence, then find the distance between two people.
Time Complexity: O(N!)

Efficient Approach: Another approach is to store person height and it’s front value, store it in any vector and sort the vector according to heights in ascending order. Now, traverse the vector and insert the person at the kth position in the position vector where k is the number of people standing in front of the people of the current person.

Below is the implementation of the above approach:

CPP

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the correct order and then 
// return the distance between the two persons 
int getDistance(int arr[][2], int n, int a, int b) 
{ 
    vector<pair<int, int> > vp; 
  
    // Make pair of both height & infront 
    // and insert to vector 
    for (int i = 0; i < n; i++) { 
        vp.push_back({ arr[i][0], arr[i][1] }); 
    } 
  
    // Sort the vector in ascending order 
    sort(vp.begin(), vp.end()); 
  
    // Find the correct place for every person 
    vector<int> pos; 
  
    // Insert into postion vector 
    // according to infront value 
    for (int i = 0; i < vp.size(); i++) { 
        int height = vp[i].first; 
        int k = vp[i].second; 
        pos.insert(pos.begin() + k, height); 
    } 
  
    int first = -1, second = -1; 
  
    for (int i = 0; i < pos.size(); i++) { 
        if (pos[i] == a) 
            first = i; 
        if (pos[i] == b) 
            second = i; 
    } 
  
    return abs(first - second); 
} 
  
// Driver code 
int main() 
{ 
    int arr[][2] = { 
        { 5, 0 }, 
        { 3, 0 }, 
        { 2, 0 }, 
        { 6, 4 }, 
        { 1, 0 }, 
        { 4, 3 } 
    }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int a = 6, b = 5; 
  
    cout << getDistance(arr, n, a, b); 
  
    return 0; 
} 

Python

# Python implementation of the approach 
  
# Function to find the correct order and then 
# return the distance between the two persons 
def getDistance(arr, n, a, b): 
    vp = [] 
  
    # Make pair of both height & infront 
    # and insert to vector 
    for i in range(n): 
        vp.append([arr[i][0], arr[i][1]]) 
  
    # Sort the vector in ascending order 
    vp = sorted(vp) 
  
    # Find the correct place for every person 
    pos = [0 for i in range(n)] 
  
    # Insert into postion vector 
    # according to infront value 
    for i in range(len(vp)): 
        height = vp[i][0] 
        k = vp[i][1] 
        pos[k] = height 
  
    first = -1
    second = -1
  
    for i in range(n): 
        if (pos[i] == a): 
            first = i 
        if (pos[i] == b): 
            second = i 
  
    return abs(first - second) 
  
# Driver code 
arr = [[5, 0], 
        [3, 0], 
        [2, 0], 
        [6, 4], 
        [1, 0], 
        [4, 3]] 
n = len(arr) 
a = 6
b = 5
  
print(getDistance(arr, n, a, b)) 
  
# This code is contributed by mohit kumar 29 

Output:

Time Complexity: O(n²)

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

CPP

Python

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