1d dp, in steps to reduce to 1

Author: dhananjay-ng

Diff for: src/main/java/problems/onRecursionAndDp/DPMinStepsToReduceToOne.java

@@ -0,0 +1,88 @@
+package problems.onRecursionAndDp;
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+public class DPMinStepsToReduceToOne {
+    /**
+     * One dimensional DP
+     * Given a number n, count minimum steps to minimise it to 1 according to the following criteria:
+     *
+     * If n is divisible by 2 then we may reduce n to n/2.
+     * If n is divisible by 3 then you may reduce n to n/3.
+     * Decrement n by 1.
+     *
+     *
+     * Input:
+     *
+     * The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains an integer N denoting the number n.
+     *
+     *
+     *
+     * Output:
+     *
+     * Output the minimum steps to minimise the number in a new line for each test case.
+     *
+     *
+     * Constraints:
+     *
+     * 1<= T <=1000
+     *
+     * 1<= N <=10000
+     *
+     *
+     * Example:
+     *
+     * Input:
+     *
+     * 2
+     *
+     * 10
+     *
+     * 6
+     *
+     * Output:
+     *
+     * 3
+     *
+     * 2
+     * @param args
+     * @throws IOException
+     */
+    public static void main (String[] args) throws IOException
+    {
+        Scanner in = new Scanner(System.in);
+        StringBuilder result = new StringBuilder();
+        int t = in.nextInt();
+        int[] nums = new int[t];
+        int max = -1;
+        for(int i=0;i<t;i++){
+            nums[i] = in.nextInt();
+            if(nums[i] > max ) max = nums[i];
+        }
+
+
+        int[] dp = new int[max+1];
+        dp[1] = 0;
+        if(max > 1)
+            dp[2] = 1;
+        if(max>2)
+            dp[3] = 1;
+
+        for(int i = 4;i<=max;i++){
+            int q1 = Integer.MAX_VALUE;
+            int q2 = Integer.MAX_VALUE;
+            int q3 = Integer.MAX_VALUE;
+            if(i % 3 == 0) q1 = 1 + dp[i / 3];
+            if(i % 2 == 0) q2 = 1 + dp[i / 2];
+            q3 = 1 + dp[i - 1];
+
+            dp[i] = Math.min(q1, Math.min(q2, q3));
+        }
+        for( int i = 0; i < t - 1; i++) {
+            result.append(dp[nums[i]]).append("\n");
+        }
+        result.append(dp[nums[t-1]]);
+        System.out.print(result.toString());
+
+    }
+}