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Minimum distance to travel to cover all intervals

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  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2022
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Given many intervals as ranges and our position. We need to find minimum distance to travel to reach such a point which covers all the intervals at once. 

Examples:  

Input : Intervals = [(0, 7), (2, 14), (4, 6)]
        Position = 3
Output : 1
We can reach position 4 by traveling
distance 1, at which all intervals will
be covered. So answer will be 1

Input : Intervals = [(1, 2), (2, 3), (3, 4)]
        Position = 2
Output : -1 
It is not possible to cover all intervals
at once at any point

Input : Intervals = [(1, 2), (2, 3), (1, 4)]
        Position = 2
Output : 0
All Intervals are covered at current 
position only so no need travel and
answer will be 0

All above examples are shown in below diagram.

We can solve this problem by concentrating only on endpoints. Since the requirement is to cover all intervals by reaching a point, all intervals must a share a point for answer to exist. Even the interval with leftmost end point must overlap with the interval right most start point. 
First, we find right most start point and left most end point from all intervals. Then we can compare our position with these points to get the result which is explained below : 

  1. If this right most start point is to the right of left most end point then it is not possible to cover all intervals simultaneously. (as in example 2)
  2. If our position is in mid between to right most start and left most end then there is no need to travel and all intervals will be covered by current position only (as in example 3)
  3. If our position is left to both points then we need to travel up to the rightmost start point and if our position is right to both points then we need to travel up to leftmost end point.

Refer above diagram to understand these cases. As in the first example, right most start is 4 and left most end is 6, so we need to reach 4 from current position 3 to cover all intervals. 

Please see the below code for a better understanding.  

C++




// C++ program to find minimum distance to
// travel to cover all intervals
#include <bits/stdc++.h>
using namespace std;
 
//  structure to store an interval
struct Interval
{
    int start, end;
    Interval(int start, int end) : start(start),
                                       end(end)
    {}
};
 
//  Method returns minimum distance to travel
// to cover all intervals
int minDistanceToCoverIntervals(Interval intervals[],
                                       int N, int x)
{
    int rightMostStart = INT_MIN;
    int leftMostEnd = INT_MAX;
 
    //  looping over all intervals to get right most
    // start and left most end
    for (int i = 0; i < N; i++)
    {
        if (rightMostStart < intervals[i].start)
            rightMostStart = intervals[i].start;
 
        if (leftMostEnd > intervals[i].end)
            leftMostEnd = intervals[i].end;
    }
     
    int res;
 
    /*  if rightmost start > leftmost end then all
        intervals are not aligned and it is not
        possible to cover all of them  */
    if (rightMostStart > leftMostEnd)
        res = -1;
 
    //  if x is in between rightmoststart and
    // leftmostend then no need to travel any distance
    else if (rightMostStart <= x && x <= leftMostEnd)
        res = 0;
     
    //  choose minimum according to current position x
    else
        res = (x < rightMostStart) ? (rightMostStart - x) :
                                     (x - leftMostEnd);
     
    return res;
}
 
//  Driver code to test above methods
int main()
{
    int x = 3;
    Interval intervals[] = {{0, 7}, {2, 14}, {4, 6}};
    int N = sizeof(intervals) / sizeof(intervals[0]);
 
    int res = minDistanceToCoverIntervals(intervals, N, x);
    if (res == -1)
        cout << "Not Possible to cover all intervals\n";
    else
        cout << res << endl;
}


Java




// Java program to find minimum distance
// to travel to cover all intervals
import java.util.*;
 
class GFG{
     
// Structure to store an interval
static class Interval
{
    int start, end;
    Interval(int start, int end)
    {
        this.start = start;
        this.end = end;
    }
};
 
// Method returns minimum distance to
// travel to cover all intervals
static int minDistanceToCoverIntervals(Interval intervals[],
                                       int N, int x)
{
    int rightMostStart = Integer.MIN_VALUE;
    int leftMostEnd = Integer.MAX_VALUE;
     
    // Looping over all intervals to get
    // right most start and left most end
    for(int i = 0; i < N; i++)
    {
        if (rightMostStart < intervals[i].start)
            rightMostStart = intervals[i].start;
 
        if (leftMostEnd > intervals[i].end)
            leftMostEnd = intervals[i].end;
    }
     
    int res;
 
    // If rightmost start > leftmost end then
    // all intervals are not aligned and it
    // is not possible to cover all of them 
    if (rightMostStart > leftMostEnd)
        res = -1;
         
    // If x is in between rightmoststart and
    // leftmostend then no need to travel
    // any distance
    else if (rightMostStart <= x &&
             x <= leftMostEnd)
        res = 0;
     
    // Choose minimum according to
    // current position x
    else
        res = (x < rightMostStart) ?
              (rightMostStart - x) :
              (x - leftMostEnd);
     
    return res;
}
 
//  Driver code
public static void main(String[] args)
{
    int x = 3;
    Interval []intervals = { new Interval(0, 7),
                             new Interval(2, 14),
                             new Interval(4, 6) };
    int N = intervals.length;
 
    int res = minDistanceToCoverIntervals(
        intervals, N, x);
     
    if (res == -1)
        System.out.print("Not Possible to " +
                         "cover all intervals\n");
    else
        System.out.print(res + "\n");
}
}
 
// This code is contributed by Rajput-Ji


Python3




# Python program to find minimum distance to
# travel to cover all intervals
 
#  Method returns minimum distance to travel
# to cover all intervals
def minDistanceToCoverIntervals(Intervals, N, x):
    rightMostStart = Intervals[0][0]
    leftMostStart = Intervals[0][1]
 
    # looping over all intervals to get right most
    # start and left most end
    for curr in Intervals:
        if rightMostStart < curr[0]:
            rightMostStart = curr[0]
        if leftMostStart > curr[1]:
            leftMostStart = curr[1]
 
    # if rightmost start > leftmost end then all
    # intervals are not aligned and it is not
    # possible to cover all of them
    if rightMostStart > leftMostStart:
        res = -1
         
    # if x is in between rightmoststart and
    # leftmostend then no need to travel any distance
    else if rightMostStart <= x and x <= leftMostStart:
        res = 0
         
    # choose minimum according to current position x
    else:
        res = rightMostStart-x if x < rightMostStart else x-leftMostStart
    return res
 
# Driver code to test above methods
Intervals = [[0, 7], [2, 14], [4, 6]]
N = len(Intervals)
x = 3
 
res = minDistanceToCoverIntervals(Intervals, N, x)
if res == -1:
    print("Not Possible to cover all intervals")
else:
    print(res)
 
    # This code is contributed by rj13to.


C#




// C# program to find minimum distance
// to travel to cover all intervals
using System;
 
class GFG{
     
// Structure to store an interval
public class Interval
{
    public int start, end;
     
    public Interval(int start, int end)
    {
        this.start = start;
        this.end = end;
    }
};
 
// Method returns minimum distance to
// travel to cover all intervals
static int minDistanceToCoverIntervals(
    Interval []intervals, int N, int x)
{
    int rightMostStart = int.MinValue;
    int leftMostEnd = int.MaxValue;
     
    // Looping over all intervals to get
    // right most start and left most end
    for(int i = 0; i < N; i++)
    {
        if (rightMostStart < intervals[i].start)
            rightMostStart = intervals[i].start;
 
        if (leftMostEnd > intervals[i].end)
            leftMostEnd = intervals[i].end;
    }
     
    int res;
 
    // If rightmost start > leftmost end then
    // all intervals are not aligned and it
    // is not possible to cover all of them 
    if (rightMostStart > leftMostEnd)
        res = -1;
         
    // If x is in between rightmoststart and
    // leftmostend then no need to travel
    // any distance
    else if (rightMostStart <= x &&
             x <= leftMostEnd)
        res = 0;
     
    // Choose minimum according to
    // current position x
    else
        res = (x < rightMostStart) ?
              (rightMostStart - x) :
              (x - leftMostEnd);
     
    return res;
}
 
//  Driver code
public static void Main(String[] args)
{
    int x = 3;
    Interval []intervals = { new Interval(0, 7),
                             new Interval(2, 14),
                             new Interval(4, 6) };
    int N = intervals.Length;
 
    int res = minDistanceToCoverIntervals(
        intervals, N, x);
     
    if (res == -1)
        Console.Write("Not Possible to " +
                      "cover all intervals\n");
    else
        Console.Write(res + "\n");
}
}
 
// This code is contributed by shikhasingrajput


Javascript




<script>
 
// JavaScript program to find minimum distance to
// travel to cover all intervals
 
// Method returns minimum distance to travel
// to cover all intervals
function minDistanceToCoverIntervals(Intervals, N, x){
    let rightMostStart = Intervals[0][0]
    let leftMostStart = Intervals[0][1]
 
    // looping over all intervals to get right most
    // start and left most end
    for(let curr of Intervals){
        if(rightMostStart < curr[0])
            rightMostStart = curr[0]
        if(leftMostStart > curr[1])
            leftMostStart = curr[1]
  }
 
  let res;
    // if rightmost start > leftmost end then all
    // intervals are not aligned and it is not
    // possible to cover all of them
    if(rightMostStart > leftMostStart)
        res = -1
         
    // if x is in between rightmoststart and
    // leftmostend then no need to travel any distance
    else if(rightMostStart <= x && x <= leftMostStart)
        res = 0
         
    // choose minimum according to current position x
    else
        res = (x < rightMostStart)?rightMostStart-x : x-leftMostStart
    return res
}
 
// Driver code to test above methods
let Intervals = [[0, 7], [2, 14], [4, 6]]
let N = Intervals.length
let x = 3
 
let res = minDistanceToCoverIntervals(Intervals, N, x)
if(res == -1)
    document.write("Not Possible to cover all intervals","<br>")
else
    document.write(res)
 
// This code is contributed by shinjanpatra
 
</script>


Output: 

1

Time Complexity: O(N)

Auxiliary Space: O(N)
This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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