CountMinSteps_iterative.py file has been created.

Author: Saket109

Python/Dynamic_Programming/CountMinSteps_Iterative.py

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+# QUESTION : 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:
+# 1.) Subtract 1 from it. (n = n - ­1) ,
+# 2.) If n is divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,
+# 3.) If n is divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ). 
+
+# Now solve this iteratively :
+
+from sys import stdin
+from sys import maxsize as MAX_VALUE
+
+
+
+def countMinStepsToOne(n,dp) :
+    dp[0] = 0
+    dp[1] = 0
+    i = 2 
+    while i<=n:
+        ans1 = dp[i-1]
+        ans2 = i-1
+        if i%2 == 0:
+            ans2 = dp[i//2]
+        ans3 = i-1
+        if i%3 == 0:
+            ans3 = dp[i//3]
+        ans = 1 + min(ans1,ans2,ans3)
+        dp[i] = ans
+        i += 1
+    return dp[n]
+
+
+#main
+n = int(stdin.readline().rstrip())
+dp = [-1 for i in range(n+1)]
+print(countMinStepsToOne(n,dp))