# Minimum cost to sort strings using reversal operations of different costs

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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] and
dp[i-1].

Computation of dp[i]
If arr[i] is greater than str[i-1], we update dp[i]
by dp[i-1]
If arr[i] is greater than reversal of previous string
we update dp[i] by dp[i-1]

Same procedure is applied to compute dp[i], we
reverse str[i] before applying the procedure.

At the end we will choose minimum of dp[N-1] and
dp[N-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 ` `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]; ` ` `  `    ``//  initializing dp array for first string ` `    ``dp = 0; ` `    ``dp = cost; ` ` `  `    ``//  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] + curCost); ` `            ``if` `(curStr >= revStr[i - 1]) ` `                ``dp[i][j] = min(dp[i][j], dp[i-1] + curCost); ` `        ``} ` `    ``} ` ` `  `    ``//  getting minimum from both entries of last index ` `    ``int` `res = min(dp[N-1], dp[N-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); ` ` `  `    ``int` `res = minCost(arr, cost, N); ` `    ``if` `(res == -1) ` `        ``cout << ``"Sorting not possible\n"``; ` `    ``else` `        ``cout << ``"Minimum cost to sort strings is "` `             ``<< res; ` `} `

## Java

 `// Java program to get minimum cost to sort ` `// strings by reversal operation ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Returns minimum cost for sorting arr[] ` `// using reverse operation. This function ` `// returns -1 if it is not possible to sort. ` `static` `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 = ``new` `int``[N][``2``]; ` ` `  `    ``// initializing dp array for first string ` `    ``dp[``0``][``0``] = ``0``; ` `    ``dp[``0``][``1``] = cost[``0``]; ` ` `  `    ``// getting array of reversed strings ` `    ``String []revStr = ``new` `String[N]; ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        ``revStr[i] = arr[i]; ` `        ``revStr[i] = reverse(revStr[i], ``0``,  ` `                            ``revStr[i].length() - ``1``); ` `    ``} ` ` `  `    ``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] = Integer.MAX_VALUE; ` ` `  `            ``// 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.compareTo(arr[i - ``1``]) >= ``0``) ` `                ``dp[i][j] = Math.min(dp[i][j],  ` `                                    ``dp[i - ``1``][``0``] + curCost); ` `            ``if` `(curStr.compareTo(revStr[i - ``1``]) >= ``0``) ` `                ``dp[i][j] = Math.min(dp[i][j], ` `                                    ``dp[i - ``1``][``1``] + curCost); ` `        ``} ` `    ``} ` ` `  `    ``// getting minimum from both entries of last index ` `    ``int` `res = Math.min(dp[N - ``1``][``0``], dp[N - ``1``][``1``]); ` ` `  `    ``return` `(res == Integer.MAX_VALUE)? -``1` `: res; ` `} ` ` `  `static` `String reverse(String s, ``int` `start, ``int` `end) ` `{ ` ` `  `    ``// Temporary variable to store character  ` `    ``char` `temp; ` `    ``char` `[]str = s.toCharArray(); ` `    ``while` `(start <= end)  ` `    ``{ ` `         `  `        ``// Swapping the first and last character  ` `        ``temp = str[start]; ` `        ``str[start] = str[end]; ` `        ``str[end] = temp; ` `        ``start++; ` `        ``end--; ` `    ``} ` `    ``return` `String.valueOf(str); ` `}  ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String arr[] = {``"aa"``, ``"ba"``, ``"ac"``}; ` `    ``int` `cost[] = {``1``, ``3``, ``1``}; ` `    ``int` `N = arr.length; ` ` `  `    ``int` `res = minCost(arr, cost, N); ` `    ``if` `(res == -``1``) ` `        ``System.out.println(``"Sorting not possible\n"``); ` `    ``else` `        ``System.out.println(``"Minimum cost to "` `+  ` `                           ``"sort strings is "` `+ res); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## 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 `

## C#

 `// C# program to get minimum cost to sort ` `// strings by reversal operation ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Returns minimum cost for sorting arr[] ` `// using reverse operation. This function ` `// returns -1 if it is not possible to sort. ` `static` `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 = ``new` `int``[N, 2]; ` ` `  `    ``// initializing dp array for first string ` `    ``dp[0, 0] = 0; ` `    ``dp[0, 1] = cost; ` ` `  `    ``// getting array of reversed strings ` `    ``String []revStr = ``new` `String[N]; ` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``{ ` `        ``revStr[i] = arr[i]; ` `        ``revStr[i] = reverse(revStr[i], 0,  ` `                            ``revStr[i].Length - 1); ` `    ``} ` ` `  `    ``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``.MaxValue; ` ` `  `            ``// 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.CompareTo(arr[i - 1]) >= 0) ` `                ``dp[i, j] = Math.Min(dp[i, j],  ` `                                    ``dp[i - 1, 0] + curCost); ` `            ``if` `(curStr.CompareTo(revStr[i - 1]) >= 0) ` `                ``dp[i, j] = Math.Min(dp[i, j], ` `                                    ``dp[i - 1, 1] + curCost); ` `        ``} ` `    ``} ` ` `  `    ``// getting minimum from both entries of last index ` `    ``int` `res = Math.Min(dp[N - 1, 0],  ` `                       ``dp[N - 1, 1]); ` ` `  `    ``return` `(res == ``int``.MaxValue) ? -1 : res; ` `} ` ` `  `static` `String reverse(String s, ``int` `start, ``int` `end) ` `{ ` ` `  `    ``// Temporary variable to store character  ` `    ``char` `temp; ` `    ``char` `[]str = s.ToCharArray(); ` `    ``while` `(start <= end)  ` `    ``{ ` `         `  `        ``// Swapping the first and last character  ` `        ``temp = str[start]; ` `        ``str[start] = str[end]; ` `        ``str[end] = temp; ` `        ``start++; ` `        ``end--; ` `    ``} ` `    ``return` `String.Join(``""``, str); ` `}  ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String []arr = {``"aa"``, ``"ba"``, ``"ac"``}; ` `    ``int` `[]cost = {1, 3, 1}; ` `    ``int` `N = arr.Length; ` ` `  `    ``int` `res = minCost(arr, cost, N); ` `    ``if` `(res == -1) ` `        ``Console.WriteLine(``"Sorting not possible\n"``); ` `    ``else` `        ``Console.WriteLine(``"Minimum cost to "` `+  ` `                          ``"sort strings is "` `+ res); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

## PHP

 `= ``\$arr``[``\$i` `- 1]) ` `                ``\$dp``[``\$i``][``\$j``] = min(``\$dp``[``\$i``][``\$j``],  ` `                                  ``\$dp``[``\$i` `- 1] +  ` `                                  ``\$curCost``); ` `            ``if` `(``\$curStr` `>= ``\$revStr``[``\$i` `- 1]) ` `                ``\$dp``[``\$i``][``\$j``] = min(``\$dp``[``\$i``][``\$j``],  ` `                                  ``\$dp``[``\$i` `- 1] +  ` `                                  ``\$curCost``); ` `        ``} ` `    ``} ` ` `  `    ``// getting minimum from both entries ` `    ``// of last index ` `    ``\$res` `= min(``\$dp``[``\$N` `- 1], ``\$dp``[``\$N` `- 1]); ` ` `  `    ``if``(``\$res` `== PHP_INT_MAX) ` `        ``return` `-1 ; ` `    ``else` `        ``return` `\$res``; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(``"aa"``, ``"ba"``, ``"ac"``); ` `\$cost` `= ``array``(1, 3, 1); ` `\$N` `= sizeof(``\$arr``); ` `\$res` `= minCost(``\$arr``, ``\$cost``, ``\$N``); ` `if` `(``\$res` `== -1) ` `    ``echo` `"Sorting not possible\n"``; ` `else` `    ``echo` `"Minimum cost to sort strings is "` `. ``\$res``; ` ` `  `// This code is contributed by ita_c ` `?> `

Output:

```Minimum cost to sort strings is 1
```

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