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DP UGUI_101
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public static int[][] table = new int[50][50]; | |
public static void main(String[] args) { | |
// write your code here | |
} | |
static void init() | |
{ | |
Arrays.fill(table,-1); | |
} | |
static int nCr(int n, int r) | |
{ | |
if(r==1) return n; | |
if (n==r) return 1; | |
if (table[n][r] != -1) | |
{ | |
return table[n][r]; | |
} | |
table[n][r] = nCr(n-1,r) +nCr(n-1,r-1); | |
return table[n][r]; | |
} |
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Problem Statement: On a positive integer, you can perform any one of the following 3 steps. | |
1.) Subtract 1 from it. ( n = n - 1 ) , | |
2.) If its divisible by 2, divide by 2. ( if n % 2 == 0 , then n = n / 2 ) , | |
3.) If its divisible by 3, divide by 3. ( if n % 3 == 0 , then n = n / 3 ). | |
Now the question is, given a positive integer n, find the minimum number of steps that takes n to 1 | |
eg: | |
1.)For n = 1 , output: 0 | |
2.) For n = 4 , output: 2 ( 4 /2 = 2 /2 = 1 ) | |
3.) For n = 7 , output: 3 ( 7 -1 = 6 /3 = 2 /2 = 1 ) |
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package com.Shubhra; | |
public class Memoization | |
{ | |
int[] memo = new int[50];// 50 is the max n | |
int getMinSteps(int n) | |
{ | |
if (n==1) return 0; | |
if (memo[n]!=-1) return memo[n]; | |
int r = 1 + getMinSteps(n-1); | |
if (n%2==0) r = Math.min(r,1+getMinSteps(n/2)); | |
if (n%3==0) r = Math.min(r,1+getMinSteps(n/3)); | |
memo[n] = r; | |
return r; | |
} | |
} |
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When you are done with the LIS, read this blog till the "Intermediate" Category, and try to solve the problems inside "Elementary" | |
Category | |
https://www.topcoder.com/community/competitive-programming/tutorials/dynamic-programming-from-novice-to-advanced/ |
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https://algorithmist.com/wiki/Longest_increasing_subsequence#Dynamic_Programming