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| 1 | +/* |
| 2 | +Given a positive integer 'n', find and return the minimum number of steps that 'n' has to take to get reduced to 1. You can perform any one of the following 3 steps: |
| 3 | +1.) Subtract 1 from it. (n = n - 1) , |
| 4 | +2.) If n is divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) , |
| 5 | +3.) If n is divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). |
| 6 | + |
| 7 | +Input format : |
| 8 | +The first and the only line of input contains an integer value, 'n'. |
| 9 | + |
| 10 | +Output format : |
| 11 | +Print the minimum number of steps. |
| 12 | + |
| 13 | +Constraints : |
| 14 | +1 <= n <= 10 ^ 6 |
| 15 | +Time Limit: 1 sec |
| 16 | + |
| 17 | +Sample Input 1 : |
| 18 | +4 |
| 19 | +Sample Output 1 : |
| 20 | +2 |
| 21 | +Explanation of Sample Output 1 : |
| 22 | +For n = 4 |
| 23 | +Step 1 : n = 4 / 2 = 2 |
| 24 | +Step 2 : n = 2 / 2 = 1 |
| 25 | + |
| 26 | +Sample Input 2 : |
| 27 | +7 |
| 28 | +Sample Output 2 : |
| 29 | +3 |
| 30 | +Explanation of Sample Output 2 : |
| 31 | +For n = 7 |
| 32 | +Step 1 : n = 7 - 1 = 6 |
| 33 | +Step 2 : n = 6 / 3 = 2 |
| 34 | +Step 3 : n = 2 / 2 = 1 |
| 35 | +*/ |
| 36 | +public class Solution { |
| 37 | + |
| 38 | + public static int countMinStepsToOne(int n) { |
| 39 | + //Your code goes here |
| 40 | + if (n==0 || n==1) |
| 41 | + return 0; |
| 42 | + else if (n==2 || n==3) |
| 43 | + return 1; |
| 44 | + |
| 45 | + int[] dp = new int[n+1]; |
| 46 | + for (int i=0;i<n+1;i++) |
| 47 | + dp[i]=-1; |
| 48 | + |
| 49 | + //Setting base cases |
| 50 | + dp[1]=0; |
| 51 | + dp[2]=1; |
| 52 | + dp[3]=1; |
| 53 | + for (int i=4;i<=n;i++) |
| 54 | + { |
| 55 | + //System.out.println("Current i: "+i); |
| 56 | + int ans1=dp[i-1]; |
| 57 | + int ans2=Integer.MAX_VALUE,ans3=Integer.MAX_VALUE; |
| 58 | + if (i%2==0) |
| 59 | + { |
| 60 | + ans2=dp[i/2]; |
| 61 | + } |
| 62 | + if (i%3==0) |
| 63 | + { |
| 64 | + ans3=dp[i/3]; |
| 65 | + } |
| 66 | + //System.out.println("ans1: "+ans1+", ans2: "+ans2+", ans3: "+ans3); |
| 67 | + dp[i]=Math.min(ans1,Math.min(ans2,ans3))+1; |
| 68 | + //System.out.println(i+": "+dp[i]); |
| 69 | + } |
| 70 | + return dp[n]; |
| 71 | + } |
| 72 | + |
| 73 | +} |
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