Consider a linear running track. Some racers are running on the track between a particular segment. A person is trying to take pictures of every racer. The person can take a picture of a racer if he stands anywhere between the racer’s running track segment i.e. between the start and end point of the racer. Given the starting and ending point of N racers and an integer D denoting the initial position of the person taking pictures. The task is to find the minimum distance the person has to move in order to take the picture of every racer from the final point. If it is impossible to take the pictures of every racer then print -1.
Examples:
Input: start[] = {0, 2, 4}, end[] = {7, 14, 6}, D = 3
Output: 1
The segments are:
0 . . . . . . 7
. . 2 . . . . . . . . . . . 14
. . . . 4 . 6
. . . d
Hence, the person has to move towards 4 i.e. 1 unit.Input: start[] = {1, 2}, end[] = {2, 3}, D = 2
Output: 0
Approach: The above problem can be solved by observation, that the person has to be ahead of each racer before they start the race and finish it. So if he is in the range of the racer starting last and the racer ending first, he can take the picture, else not.
Find the maximum value of the starting point say left and the minimum value of the ending point say right among all the given racers. Now,
- left > right then it is impossible for the person to take pictures and print -1.
- If D is within the range [left, right] then the person doesn’t need to move and the answer will be 0.
- Else the person has to move min(abs(left – D), abs(right – D)).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum// distance the person has to move// int order to get the picturesint minDistance(int start[], int end[], int n, int d){ // To store the minimum ending point int left = INT_MIN; // To store the maximum starting point int right = INT_MAX; // Find the values of minSeg and maxSeg for (int i = 0; i < n; i++) { left = max(left, start[i]); right = min(right, end[i]); } // Impossible if (left > right) return -1; // The person doesn't need to move if (d >= left && d <= right) return 0; // Get closer to the left point if (d < left) return (left - d); // Get closer to the right point if (d > right) return (d - right);}// Driver codeint main(){ int start[] = { 0, 2, 4 }; int end[] = { 7, 14, 6 }; int n = sizeof(start) / sizeof(int); int d = 3; cout << minDistance(start, end, n, d); return 0;} |
Java
// Java implementation for the above approachimport java.io.*;class GFG{ // Function to return the minimum // distance the person has to move // int order to get the pictures static int minDistance(int start[], int end[], int n, int d) { // To store the minimum ending point int left = Integer.MIN_VALUE; // To store the maximum starting point int right = Integer.MAX_VALUE; // Find the values of minSeg and maxSeg for (int i = 0; i < n; i++) { left = Math.max(left, start[i]); right = Math.min(right, end[i]); } // Impossible if (left > right) return -1; // The person doesn't need to move if (d >= left && d <= right) return 0; // Get closer to the left point if (d < left) return (left - d); // Get closer to the right point if (d > right) return (d - right); return -1; } // Driver code public static void main(String[] args) { int start[] = { 0, 2, 4 }; int end[] = { 7, 14, 6 }; int n = start.length; int d = 3; System.out.println(minDistance(start, end, n, d)); }}// This code is contributed by Potta Lokesh |
Python3
# Python3 program for the above approachimport sys# Function to return the minimum# distance the person has to move# order to get the picturesdef minDistance(start, end, n, d) : # To store the minimum ending point left = -sys.maxsize # To store the maximum starting point right = sys.maxsize # Find the values of minSeg and maxSeg for i in range(n) : left = max(left, start[i]) right = min(right, end[i]) # Impossible if (left > right): return -1 # The person doesn't need to move if (d >= left and d <= right): return 0 # Get closer to the left point if (d < left) : return (left - d) # Get closer to the right point if (d > right) : return (d - right)# Driver codestart = [ 0, 2, 4 ] end = [ 7, 14, 6 ]n = len(start)d = 3print(minDistance(start, end, n, d))# This code is contributed by target_2. |
C#
// C# program for above approachusing System;class GFG{// Function to return the minimum// distance the person has to move// int order to get the picturesstatic int minDistance(int[] start, int[] end, int n, int d){ // To store the minimum ending point int left = Int32.MinValue; // To store the maximum starting point int right = Int32.MaxValue; // Find the values of minSeg and maxSeg for(int i = 0; i < n; i++) { left = Math.Max(left, start[i]); right = Math.Min(right, end[i]); } // Impossible if (left > right) return -1; // The person doesn't need to move if (d >= left && d <= right) return 0; // Get closer to the left point if (d < left) return (left - d); // Get closer to the right point if (d > right) return (d - right); return -1;}// Driver Codepublic static void Main(String[] args){ int[] start = { 0, 2, 4 }; int[] end = { 7, 14, 6 }; int n = start.Length; int d = 3; Console.Write(minDistance(start, end, n, d));}}// This code is contributed by sanjoy_62 |
Javascript
<script>// JavaScript implementation of the approach// Function to return the minimum// distance the person has to move// int order to get the picturesfunction minDistance(start, intend, n, d) { // To store the minimum ending point let left = Number.MIN_SAFE_INTEGER; // To store the maximum starting point let right = Number.MAX_SAFE_INTEGER; // Find the values of minSeg and maxSeg for (let i = 0; i < n; i++) { left = Math.max(left, start[i]); right = Math.min(right, end[i]); } // Impossible if (left > right) return -1; // The person doesn't need to move if (d >= left && d <= right) return 0; // Get closer to the left point if (d < left) return left - d; // Get closer to the right point if (d > right) return d - right;}// Driver codelet start = [0, 2, 4];let end = [7, 14, 6];let n = start.length;let d = 3;document.write(minDistance(start, end, n, d));</script> |
1
Time complexity: O(N) where N is the number of racers
Auxiliary space: O(1)
