모든 간격을 커버하기 위한 최소 이동 거리
범위와 위치로 많은 간격이 주어집니다. 모든 구간을 동시에 포괄하는 지점에 도달하려면 이동해야 할 최소 거리를 찾아야 합니다.
예:
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.
엔드포인트에만 집중하면 이 문제를 해결할 수 있습니다. 요구 사항은 한 지점에 도달하여 모든 간격을 포괄하는 것이므로 답변이 존재하려면 모든 간격이 한 지점을 공유해야 합니다. 가장 왼쪽 끝점이 있는 구간도 가장 오른쪽 시작점이 있는 구간과 겹쳐야 합니다.
먼저 모든 간격에서 가장 오른쪽 시작점과 가장 왼쪽 끝점을 찾습니다. 그런 다음 우리의 위치를 이러한 지점과 비교하여 아래에 설명된 결과를 얻을 수 있습니다.
- 이 가장 오른쪽 시작점이 가장 왼쪽 끝점의 오른쪽에 있으면 모든 간격을 동시에 포함할 수 없습니다. (예 2에서와 같이)
- 우리 위치가 가장 오른쪽 시작과 가장 왼쪽 끝 사이의 중간에 있으면 이동할 필요가 없으며 모든 간격은 현재 위치로만 처리됩니다(예 3에서와 같이).
- 우리 위치가 두 지점 모두에 남아 있으면 가장 오른쪽 시작 지점까지 이동해야 하고, 위치가 두 지점 모두 오른쪽에 있으면 가장 왼쪽 끝 지점까지 이동해야 합니다.
이러한 경우를 이해하려면 위의 다이어그램을 참조하세요. 첫 번째 예에서와 같이 가장 오른쪽 시작은 4이고 왼쪽 가장 끝은 6이므로 모든 간격을 포괄하려면 현재 위치 3에서 4에 도달해야 합니다.
더 나은 이해를 위해 아래 코드를 참조하십시오.
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 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 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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