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すべての間隔をカバーするための最小移動距離

範囲と私たちの位置として多くの間隔が与えられます。すべての区間を一度にカバーするような地点に到達するための最小移動距離を見つける必要があります。 

例:   

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.

すべての間隔をカバーするための最小移動距離



この問題は、エンドポイントのみに集中することで解決できます。要件は、ポイントに到達することですべての間隔をカバーすることであるため、答えが存在するには、すべての間隔がポイントを共有する必要があります。左端の終点を持つ区間であっても、区間の右端の始点と重なる必要があります。 
まず、すべての間隔から右端の開始点と左端の終了点を見つけます。次に、自分の位置をこれらのポイントと比較して、以下で説明する結果を得ることができます。 

  1. この右端の開始点が左端の終了点の右側にある場合、すべての間隔を同時にカバーすることはできません。 (例 2 と同様)
  2. 位置が右端の開始点と左端の端点の間の中間にある場合、移動する必要はなく、すべての間隔は現在の位置のみでカバーされます (例 3 のように)
  3. 私たちの位置が両方の点に対して左にある場合は、右端の開始点まで移動する必要があり、私たちの位置が両方の点に対して右である場合は、左端の終点まで移動する必要があります。

これらのケースを理解するには、上の図を参照してください。最初の例と同様に、右端の開始は 4、左端は 6 であるため、すべての間隔をカバーするには、現在の位置 3 から 4 に到達する必要があります。 

よりよく理解するには、以下のコードを参照してください。  

C++ 
// C++ program to find minimum distance to  // travel to cover all intervals #include    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 intervalsn';  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 intervalsn');  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 intervalsn');  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''  
') else  document.write(res) // This code is contributed by shinjanpatra </script>

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