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Create Stair Case

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Course 2 - Data Structures in JAVA/Lecture 18 - Dynamic Programming I
- Stair Case

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	`1`	`+/*`
	`2`	`+A child runs up a staircase with 'n' steps and can hop either 1 step, 2 steps or 3 steps at a time.`
	`3`	`+Implement a method to count and return all possible ways in which the child can run-up to the stairs.`
	`4`	`+`
	`5`	`+Input format :`
	`6`	`+The first and the only line of input contains an integer value, 'n', denoting the total number of steps.`
	`7`	`+`
	`8`	`+Output format :`
	`9`	`+Print the total possible number of ways.`
	`10`	`+`
	`11`	`+Constraints :`
	`12`	`+0 <= n <= 10 ^ 4`
	`13`	`+Time Limit: 1 sec`
	`14`	`+`
	`15`	`+Sample Input 1:`
	`16`	`+4`
	`17`	`+Sample Output 1:`
	`18`	`+7`
	`19`	`+`
	`20`	`+Sample Input 2:`
	`21`	`+10`
	`22`	`+Sample Output 2:`
	`23`	`+274`
	`24`	`+*/`
	`25`	`+public class Solution {`
	`26`	`+`
	`27`	`+ public static long staircase(int n) {`
	`28`	`+ //Your code goes here`
	`29`	`+ /*`
	`30`	`+ //If we reach n=0, we have found a path`
	`31`	`+ if (n==0)`
	`32`	`+ return 1;`
	`33`	`+`
	`34`	`+ //If n<0, the steps we took till then don't work`
	`35`	`+ else if(n<0)`
	`36`	`+ return 0;`
	`37`	`+`
	`38`	`+ return staircase(n-1)+staircase(n-2)+staircase(n-3);`
	`39`	`+ */`
	`40`	`+`
	`41`	`+ if(n<=1)`
	`42`	`+ return 1;`
	`43`	`+ if(n==2)`
	`44`	`+ return 2;`
	`45`	`+`
	`46`	`+ long dpCount[] = new long[n+1];`
	`47`	`+ dpCount[0]=1;`
	`48`	`+ dpCount[1]=1;`
	`49`	`+ dpCount[2]=2;`
	`50`	`+`
	`51`	`+`
	`52`	`+ for (int i=3;i<=n;i++)`
	`53`	`+ {`
	`54`	`+ dpCount[i]=dpCount[i-1]+dpCount[i-2]+dpCount[i-3];`
	`55`	`+`
	`56`	`+ }`
	`57`	`+ return dpCount[n];`
	`58`	`+ }`
	`59`	`+`
	`60`	`+}`

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