Number of jumps for a thief to cross walls
A thief trying to escape from a jail. He has to cross N walls each with varying heights (every height is greater than 0). He climbs X feet every time. But, due to the slippery nature of those walls, every time he slips back by Y feet. Now the task is to calculate the total number of jumps required to cross all walls and escape from the jail.
Examples :
Input : heights[] = {11, 11} X = 10; Y = 1; Output : 4 He needs to make 2 jumps for first wall and 2 jumps for second wall. Input : heights[] = {11, 10, 10, 9} X = 10; Y = 1; Output : 5
The solution is quite simple if the height of wall is less than or equal to x, only one jump in that wall is required else we can calculate it by height of wall-(climb up-climb down) and get the jumps required.
C++
// C++ program to find the number of // jumps required #include <iostream> using namespace std; /* function to calculate jumps required to cross walls */ int jumpcount( int x, int y, int n, int height[]) { int jumps = 0; for ( int i = 0; i < n; i++) { if (height[i] <= x) { jumps++; continue ; } /* If height of wall is greater than up move */ int h = height[i]; while (h > x) { jumps++; h = h - (x - y); } jumps++; } return jumps; } /*driver function*/ int main() { int x = 10, y = 1; int height[] = { 11, 10, 10, 9 }; int n = sizeof (height)/ sizeof (height[0]); cout << jumpcount(x, y, n, height); return 0; } |
Java
// Java program to find the number of // jumps required public class GFG { /* function to calculate jumps required to cross walls */ static int jumpcount( int x, int y, int n, int height[]) { int jumps = 0 ; for ( int i = 0 ; i < n; i++) { if (height[i] <= x) { jumps++; continue ; } /* If height of wall is greater than up move */ int h = height[i]; while (h > x) { jumps++; h = h - (x - y); } jumps++; } return jumps; } /*driver function*/ public static void main(String args[]) { int x = 10 , y = 1 ; int height[] = { 11 , 10 , 10 , 9 }; int n = height.length; System.out.println(jumpcount(x, y, n, height)); } } // This code is contributed by Sumit Ghosh |
Python3
# Python program to find the number of # jumps required # function to calculate jumps required to # cross walls def jumpcount(x, y, n, height): jumps = 0 for i in range (n): if (height[i] < = x): jumps + = 1 continue """ If height of wall is greater than up move """ h = height[i] while (h > x): jumps + = 1 h = h - (x - y) jumps + = 1 return jumps # driver function x = 10 y = 1 height = [ 11 , 10 , 10 , 9 ] n = len (height) print (jumpcount(x, y, n, height)) # This code is contributed by Sachin Bisht |
C#
// C# program to find the // number of jumps required using System; public class GFG { // function to calculate jumps // required to cross walls static int jumpcount( int x, int y, int n, int []height) { int jumps = 0; for ( int i = 0; i < n; i++) { if (height[i] <= x) { jumps++; continue ; } // If height of wall is // greater than up move int h = height[i]; while (h > x) { jumps++; h = h - (x - y); } jumps++; } return jumps; } // Driver Code public static void Main(String []args) { int x = 10, y = 1; int []height = {11, 10, 10, 9}; int n = height.Length; Console.WriteLine(jumpcount(x, y, n, height)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find the // number of jumps required /* function to calculate jumps required to cross walls */ function jumpcount( $x , $y , $n , $height ) { $jumps = 0; for ( $i = 0; $i < $n ; $i ++) { if ( $height [ $i ] <= $x ) { $jumps ++; continue ; } /* If height of wall is greater than up move */ $h = $height [ $i ]; while ( $h > $x ) { $jumps ++; $h = $h - ( $x - $y ); } $jumps ++; } return $jumps ; } // Driver Code $x = 10; $y = 1; $height = array ( 11, 10, 10, 9 ); $n = count ( $height ); echo jumpcount( $x , $y , $n , $height ); // This code is contributed by anuj_67. ?> |
Javascript
<script> // JavaScript program to find the // number of jumps required // function to calculate jumps // required to cross walls function jumpcount(x, y, n, height) { let jumps = 0; for (let i = 0; i < n; i++) { if (height[i] <= x) { jumps++; continue ; } // If height of wall is // greater than up move let h = height[i]; while (h > x) { jumps++; h = h - (x - y); } jumps++; } return jumps; } let x = 10, y = 1; let height = [11, 10, 10, 9]; let n = height.length; document.write(jumpcount(x, y, n, height)); </script> |
Output:
5
Time Complexity: O(height[i]*n)
Auxiliary Space: O(1)
We can optimize above solution by directly computing number of jumps.
C++
// C++ program to find the number of // jumps required #include <iostream> using namespace std; /* function to calculate jumps required to cross walls */ int jumpcount( int x, int y, int n, int height[]) { int jumps = 0; for ( int i = 0; i < n; i++) { // Since all heights are // greater than 1, at-least // one jump is always required jumps++; // More jumps required if height // is greater than x. if (height[i] > x) { // Since we have already counted // one jump int h = height[i] - (x - y); // Remaining jumps jumps += h/(x - y); // If there was a remainder greater // than 1. 1 is there to handle cases // like x = 11, y = 1, height[i] = 21. if (h % (x-y) > 1) jumps++; } } return jumps; } /* driver function */ int main() { int x = 10; int y = 1; int height[] = { 11, 34, 27, 9 }; int n = sizeof (height)/ sizeof (height[0]); cout << jumpcount(x, y, n, height); return 0; } |
Java
// Java program to find the number of // jumps required public class GFG { /* function to calculate jumps required to cross walls */ static int jumpcount( int x, int y, int n, int height[]) { int jumps = 0 ; for ( int i = 0 ; i < n; i++) { // Since all heights are // greater than 1, at-least // one jump is always required jumps++; // More jumps required if height // is greater than x. if (height[i] > x) { // Since we have already counted // one jump int h = height[i] - (x - y); // Remaining jumps jumps += h/(x - y); // If there was a remainder greater // than 1. 1 is there to handle cases // like x = 11, y = 1, height[i] = 21. if (h % (x-y) > 1 ) jumps++; } } return jumps; } /* driver function */ public static void main(String args[]) { int x = 10 ; int y = 1 ; int height[] = { 11 , 34 , 27 , 9 }; int n = height.length; System.out.println(jumpcount(x, y, n, height)); } } // This code is contributed by Sumit Ghosh |
Python3
# Python program to find the number of # jumps required """ function to calculate jumps required to cross walls """ def jumpcount(x, y, n, height): jumps = 0 for i in range (n): # Since all heights are # greater than 1, at-least # one jump is always required jumps + = 1 # More jumps required if height # is greater than x. if (height[i] > x): # Since we have already counted # one jump h = height[i] - (x - y) # Remaining jumps jumps + = h / / (x - y) # If there was a remainder greater # than 1. 1 is there to handle cases # like x = 11, y = 1, height[i] = 21. if (h % (x - y) > 1 ): jumps + = 1 return jumps # driver function x = 10 y = 1 height = [ 11 , 34 , 27 , 9 ] n = len (height) print (jumpcount(x, y, n, height)) # This code is contributed by Sachin Bisht |
C#
// C# program to find the // number of jumps required using System; public class GFG { // Function to calculate jumps // required to cross walls static int jumpcount( int x, int y, int n, int []height) { int jumps = 0; for ( int i = 0; i < n; i++) { // Since all heights are // greater than 1, at-least // one jump is always required jumps++; // More jumps required if // height is greater than x. if (height[i] > x) { // Since we have already // counted one jump int h = height[i] - (x - y); // Remaining jumps jumps += h / (x - y); // If there was a remainder greater // than 1. 1 is there to handle cases // like x = 11, y = 1, height[i] = 21. if (h % (x - y) > 1) jumps++; } } return jumps; } // Driver Code public static void Main(String []args) { int x = 10; int y = 1; int []height = {11, 34, 27, 9}; int n = height.Length; Console.WriteLine(jumpcount(x, y, n, height)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find the // number of jumps required // function to calculate jumps // required to cross walls function jumpcount( $x , $y , $n , $height ) { $jumps = 0; for ( $i = 0; $i < $n ; $i ++) { // Since all heights are // greater than 1, at-least // one jump is always required $jumps ++; // More jumps required if height // is greater than x. if ( $height [ $i ] > $x ) { // Since we have // already counted // one jump $h = $height [ $i ] - ( $x - $y ); // Remaining jumps $jumps += $h / ( $x - $y ); // If there was a // remainder greater // than 1. 1 is there // to handle cases // like x = 11, y = 1, // height[i] = 21. if ( $h % ( $x - $y ) > 1) $jumps ++; } } return $jumps ; } // Driver Code { $x = 10; $y = 1; $height = array (11, 34, 27, 9); $n = sizeof( $height ) / sizeof( $height [0]); echo jumpcount( $x , $y , $n , $height ); return 0; } // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript program to find the number of jumps required // Function to calculate jumps // required to cross walls function jumpcount(x, y, n, height) { let jumps = 0; for (let i = 0; i < n; i++) { // Since all heights are // greater than 1, at-least // one jump is always required jumps++; // More jumps required if // height is greater than x. if (height[i] > x) { // Since we have already // counted one jump let h = height[i] - (x - y); // Remaining jumps jumps += parseInt(h / (x - y), 10); // If there was a remainder greater // than 1. 1 is there to handle cases // like x = 11, y = 1, height[i] = 21. if (h % (x - y) > 1) jumps++; } } return jumps; } let x = 10; let y = 1; let height = [11, 34, 27, 9]; let n = height.length; document.write(jumpcount(x, y, n, height)); // This code is contributed by mukesh07. </script> |
Output:
10
Time Complexity: O(n)
Auxiliary Space: O(1)