Doti daudzi intervāli kā diapazoni un mūsu pozīcija. Mums ir jāatrod minimālais attālums, lai sasniegtu punktu, kas aptver visus intervālus vienlaikus.
Piemēri:
Input : Intervals = [(0 7) (2 14) (4 6)] Position = 3 Output : 1 We can reach position 4 by travelling 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.
Mēs varam atrisināt šo problēmu, koncentrējoties tikai uz galapunktiem. Tā kā prasība ir aptvert visus intervālus, sasniedzot punktu, lai atbilde pastāvētu, visiem intervāliem ir jābūt vienam punktam. Pat intervālam ar galējo beigu punktu pa kreisi jāpārklājas ar intervāla sākuma punktu, kas atrodas vistālāk pa kreisi.
Vispirms no visiem intervāliem atrodam labāko sākuma punktu un kreiso beigu punktu. Tad mēs varam salīdzināt savu pozīciju ar šiem punktiem, lai iegūtu rezultātu, kas izskaidrots zemāk:
- Ja šis sākuma punkts atrodas pa labi no kreisā gala punkta, tad nav iespējams aptvert visus intervālus vienlaicīgi. (kā 2. piemērā)
- Ja mūsu pozīcija ir vidū starp lielāko daļu no labās puses uz lielāko daļu un no kreisās malas gala, tad nav nepieciešams ceļot un visus intervālus sedz tikai pašreizējā pozīcija (kā 3. piemērā).
- Ja mūsu pozīcija ir pa kreisi abos punktos, mums jādodas līdz galējam labajam sākuma punktam, un, ja mūsu pozīcija ir pa labi uz abiem punktiem, tad mums jābrauc līdz galējam kreisajam beigu punktam.
Lai izprastu šos gadījumus, skatiet iepriekš minēto diagrammu. Tāpat kā pirmajā piemērā labās daļas sākums ir 4 un kreisās puses gals ir 6, tāpēc mums ir jāsasniedz 4 no pašreizējās pozīcijas 3, lai aptvertu visus intervālus.
Lai labāk izprastu, lūdzu, skatiet tālāk norādīto kodu.
C++// C++ program to find minimum distance to // travel to cover all intervals #include
// 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 intervalsn'); else System.out.print(res + 'n'); } } // This code is contributed by Rajput-Ji
# 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# 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 intervalsn'); else Console.Write(res + 'n'); } } // This code is contributed by shikhasingrajput
<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''
') else document.write(res) // This code is contributed by shinjanpatra </script>
Izvade:
1
Laika sarežģītība: O(N)
Palīgtelpa: O(N)