Solution 4 for Scaler Topics Fortnightly Contest - 2

This article is part of the Scaler Topics Fortnightly Contest - 2.

Solution Approach

• You can keep a 2D dp, where first parameter will be the index of string and second parameter will be whether you have reversed this string or not. Initialize your dp to a large number.
• Let Br be the array where all the strings are reversed.
• Now for each index i check these four cases : if B[i-1] <= B[i] , then update your dp[i][0] to min of dp[i][0], dp[i-1][0]. if Br[i-1] <= B[i] , then update your dp[i][0] to min of dp[i][0], dp[i-1][1]. if B[i-1] <= Br[i] , then update your dp[i][1] to min of dp[i][1], dp[i-1][0]. if Br[i-1] <= Br[i] , then update your dp[i][1] to min of dp[i][1], dp[i-1][1]. And your answer will be minimum of dp[n][0] and dp[n][1].
• If the answer is equal to or greater thant the large number to initialized your dp with. Then it means you cannot sort your array.

C++ Implementation

Java Implementation

Python Implementation