Najmanjša potovalna razdalja za pokrivanje vseh intervalov
Glede na številne intervale kot razpone in naš položaj. Najti moramo najmanjšo razdaljo, ki jo moramo prepotovati, da dosežemo takšno točko, ki pokriva vse intervale hkrati.
Primeri:
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.
To težavo lahko rešimo tako, da se osredotočimo le na končne točke. Ker je zahteva pokriti vse intervale z doseganjem točke, morajo imeti vsi intervali skupno točko, da obstaja odgovor. Tudi interval s skrajno levo končno točko se mora prekrivati s skrajno desno začetno točko intervala.
Najprej poiščemo skrajno desno začetno točko in skrajno levo končno točko iz vseh intervalov. Nato lahko primerjamo svoj položaj s temi točkami, da dobimo rezultat, ki je razložen spodaj:
- Če je ta skrajna desna začetna točka desno od skrajne leve končne točke, potem ni mogoče pokriti vseh intervalov hkrati. (kot v primeru 2)
- Če je naš položaj na sredini med skrajnim desnim začetkom in skrajnim levim koncem, ni potrebe po potovanju in vsi intervali bodo zajeti samo s trenutnim položajem (kot v primeru 3)
- Če je naš položaj levo do obeh točk, moramo potovati do skrajno desne začetne točke in če je naš položaj desno do obeh točk, potem moramo potovati do skrajno leve končne točke.
Za razumevanje teh primerov glejte zgornji diagram. Kot v prvem primeru je skrajni desni začetek 4 in skrajni levi konec 6, zato moramo doseči 4 s trenutnega položaja 3, da pokrijemo vse intervale.
Za boljše razumevanje si oglejte spodnjo kodo.
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>
Izhod:
1
Časovna zapletenost: O(N)
Pomožni prostor: O(N)
