Given an array of strings and costs of reversing all strings, we need to sort the array. We cannot move strings in array, only string reversal is allowed. We need to reverse some of the strings in such a way that all strings make a lexicographic order and cost is also minimized. If it is not possible to sort strings in any way, output not possible.

Examples:

Input : arr[] = {“aa”, “ba”, “ac”}, reverseCost[] = {1, 3, 1} Output : Minimum cost of sorting = 1 Explanation : We can make above string array sorted by reversing one of 2nd or 3rd string, but reversing 2nd string cost 3, so we will reverse 3rd string to make string array sorted with a cost 1 which is minimum.

We can solve this problem using dynamic programming. We make a 2D array for storing the minimum cost of sorting.

dp[i][j] represents the minimum cost to make first i strings sorted. j = 1 means i'th string is reversed. j = 0 means i'th string is not reversed. Value of dp[i][j] is computed using dp[i-1][1] and dp[i-1][0]. Computation ofdp[i][0]If arr[i] is greater than str[i-1], we update dp[i][0] by dp[i-1][0] If arr[i] is greater than reversal of previous string we update dp[i][0] by dp[i-1][1] Same procedure is applied to computedp[i][1], we reverse str[i] before applying the procedure. At the end we will choose minimum of dp[N-1][0] and dp[N-1][1] as our final answer if both of them not updated yet even once, we will flag that sorting is not possible.

Below is the implementation of above idea.

## C/C++

// C++ program to get minimum cost to sort // strings by reversal operation #include <bits/stdc++.h> using namespace std; // Returns minimum cost for sorting arr[] // using reverse operation. This function // returns -1 if it is not possible to sort. int minCost(string arr[], int cost[], int N) { // dp[i][j] represents the minimum cost to // make first i strings sorted. // j = 1 means i'th string is reversed. // j = 0 means i'th string is not reversed. int dp[N][2]; // initializing dp array for first string dp[0][0] = 0; dp[0][1] = cost[0]; // getting array of reversed strings string revStr[N]; for (int i = 0; i < N; i++) { revStr[i] = arr[i]; reverse(revStr[i].begin(), revStr[i].end()); } string curStr; int curCost; // looping for all strings for (int i = 1; i < N; i++) { // Looping twice, once for string and once // for reversed string for (int j = 0; j < 2; j++) { dp[i][j] = INT_MAX; // getting current string and current // cost according to j curStr = (j == 0) ? arr[i] : revStr[i]; curCost = (j == 0) ? 0 : cost[i]; // Update dp value only if current string // is lexicographically larger if (curStr >= arr[i - 1]) dp[i][j] = min(dp[i][j], dp[i-1][0] + curCost); if (curStr >= revStr[i - 1]) dp[i][j] = min(dp[i][j], dp[i-1][1] + curCost); } } // getting minimum from both entries of last index int res = min(dp[N-1][0], dp[N-1][1]); return (res == INT_MAX)? -1 : res; } // Driver code to test above methods int main() { string arr[] = {"aa", "ba", "ac"}; int cost[] = {1, 3, 1}; int N = sizeof(arr) / sizeof(arr[0]); int res = minCost(arr, cost, N); if (res == -1) cout << "Sorting not possible\n"; else cout << "Minimum cost to sort strings is " << res; }

## Python

# Python program to get minimum cost to sort # strings by reversal operation # Returns minimum cost for sorting arr[] # using reverse operation. This function # returns -1 if it is not possible to sort. def ReverseStringMin(arr, reverseCost, n): # dp[i][j] represents the minimum cost to # make first i strings sorted. # j = 1 means i'th string is reversed. # j = 0 means i'th string is not reversed. dp = [[float("Inf")] * 2 for i in range(n)] # initializing dp array for first string dp[0][0] = 0 dp[0][1] = reverseCost[0] # getting array of reversed strings rev_arr = [i[::-1] for i in arr] # looping for all strings for i in range(1, n): # Looping twice, once for string and once # for reversed string for j in range(2): # getting current string and current # cost according to j curStr = arr[i] if j==0 else rev_arr[i] curCost = 0 if j==0 else reverseCost[i] # Update dp value only if current string # is lexicographically larger if (curStr >= arr[i - 1]): dp[i][j] = min(dp[i][j], dp[i-1][0] + curCost) if (curStr >= rev_arr[i - 1]): dp[i][j] = min(dp[i][j], dp[i-1][1] + curCost) # getting minimum from both entries of last index res = min(dp[n-1][0], dp[n-1][1]) return res if res != float("Inf") else -1 # Driver code def main(): arr = ["aa", "ba", "ac"] reverseCost = [1, 3, 1] n = len(arr) dp = [float("Inf")] * n res = ReverseStringMin(arr, reverseCost,n) if res != -1 : print "Minimum cost to sort sorting is" , res else : print "Sorting not possible" if __name__ == '__main__': main() #This code is contributed by Neelam Yadav

Output:

Minimum cost to sort strings is 1

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