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
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:
// 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 position 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;
} |
/*package whatever //do not write package name here */ import java.util.*;
class GFG {
// Function to find the correct order and then
// return the distance between the two persons
static int getDistance( int [][] arr, int n, int a, int b)
{
int vp[][] = new int [n][ 2 ];
// Make pair of both height & infront
// and insert to vector
for ( int i= 0 ; i < n; i++) {
vp[i][ 0 ] = arr[i][ 0 ];
vp[i][ 1 ] = arr[i][ 1 ];
}
// Sort the vector in ascending order
Arrays.sort(vp, (c,d)-> c[ 0 ]-d[ 0 ]);
// Find the correct place for every person
int pos[] = new int [n];
for ( int i = 0 ; i < n; i++) {
pos[i] = 0 ;
}
// Insert into position vector
// according to infront value
for ( int i = 0 ; i < vp.length; i++) {
int height = vp[i][ 0 ];
int k = vp[i][ 1 ];
pos[k] = height;
}
int first = - 1 , second = - 1 ;
for ( int i = 0 ; i < pos.length; i++) {
if (pos[i] == a)
first = i;
if (pos[i] == b)
second = i;
}
return Math.abs(first - second);
}
public static void main(String[] args)
{
// Driver code
int arr[][] = { { 5 , 0 }, { 3 , 0 }, { 2 , 0 },
{ 6 , 4 }, { 1 , 0 }, { 4 , 3 } };
int n = arr.length;
int a = 6 , b = 5 ;
System.out.println(getDistance(arr, n, a, b));
}
} // This code is contributed by aadityaburujwale. |
# 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 position 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 |
using System;
using System.Collections.Generic;
public class GFG {
public static int cmp(KeyValuePair< int , int > a,
KeyValuePair< int , int > b)
{
if (a.Key > b.Key)
return 1;
return -1;
}
// Function to find the correct order and then
// return the distance between the two persons
public static int getDistance( int [, ] arr, int n, int a,
int b)
{
var vp = new List<KeyValuePair< int , int > >();
// Make pair of both height & infront
// and insert to vector
for ( int i = 0; i < n; i++) {
vp.Add( new KeyValuePair< int , int >(arr[i, 0],
arr[i, 1]));
}
// Sort the vector in ascending order
vp.Sort(cmp);
// Find the correct place for every person
List< int > pos = new List< int >();
for ( int i = 0; i < 100; i++) {
pos.Add(0);
}
// Insert into position vector
// according to infront value
for ( int i = 0; i < vp.Count; i++) {
int height = vp[i].Key;
int k = vp[i].Value;
pos[k] = height;
}
int first = -1, second = -1;
for ( int i = 0; i < pos.Count; i++) {
if (pos[i] == a)
first = i;
if (pos[i] == b)
second = i;
}
return Math.Abs(first - second);
}
static public void Main()
{
int [, ] arr = { { 5, 0 }, { 3, 0 }, { 2, 0 },
{ 6, 4 }, { 1, 0 }, { 4, 3 } };
int n = arr.GetLength(0);
int a = 6, b = 5;
Console.WriteLine(getDistance(arr, n, a, b));
}
} // This code is contributed by akashish__ |
<script> // Javascript implementation of the approach // Function to find the correct order and then // return the distance between the two persons function getDistance(arr,n,a,b)
{ let vp=[];
// Make pair of both height & infront
// and insert to vector
for (let i = 0; i < n; i++) {
vp.push([ arr[i][0], arr[i][1] ]);
}
// Sort the vector in ascending order
vp.sort( function (c,d){ return c[0]-d[0];});
// Find the correct place for every person
let pos= new Array(n);
for (let i=0;i<n;i++)
{
pos[i]=0;
}
// Insert into position vector
// according to infront value
for (let i = 0; i < vp.length; i++) {
let height = vp[i][0];
let k = vp[i][1];
pos[k] = height
}
let first = -1, second = -1;
for (let i = 0; i < pos.length; i++) {
if (pos[i] == a)
first = i;
if (pos[i] == b)
second = i;
}
return Math.abs(first - second);
} // Driver code let arr = [ [ 5, 0 ],
[ 3, 0 ],
[ 2, 0 ],
[ 6, 4 ],
[ 1, 0 ],
[ 4, 3 ]
];
let n = arr.length;
let a = 6, b = 5;
document.write(getDistance(arr, n, a, b));
// This code is contributed by unknown2108 </script> |
4
Time Complexity: O(n2)
Space Complexity: O(n) as arrays like pos and vp has been created. Here, n is the size of the input array.