# 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.

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++ 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;` `}` |

## 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` |

## Python3

`# 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[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] = ` `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

`<?php ` `// PHP 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.` `function` `minCost(&` `$arr` `, &` `$cost` `, ` `$N` `)` `{` ` ` `// dp[i][j] represents the minimum cost ` ` ` `// to make first i strings sorted.` ` ` `// j = 1 means i'th string is reversed.` ` ` `// j = 0 means i'th string is not reversed.` ` ` `$dp` `= ` `array_fill` `(0, ` `$N` `, ` ` ` `array_fill` `(0, 2, NULL));` ` ` `// initializing dp array for ` ` ` `// first string` ` ` `$dp` `[0][0] = 0;` ` ` `$dp` `[0][1] = ` `$cost` `[0];` ` ` `// getting array of reversed strings` ` ` `$revStr` `= ` `array_fill` `(false, ` `$N` `, NULL);` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$N` `; ` `$i` `++)` ` ` `{` ` ` `$revStr` `[` `$i` `] = ` `$arr` `[` `$i` `];` ` ` `$revStr` `[` `$i` `] = ` `strrev` `(` `$revStr` `[` `$i` `]);` ` ` `}` ` ` `$curStr` `= ` `""` `;` ` ` `// looping for all strings` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$N` `; ` `$i` `++)` ` ` `{` ` ` `// Looping twice, once for string ` ` ` `// and once for reversed string` ` ` `for` `(` `$j` `= 0; ` `$j` `< 2; ` `$j` `++)` ` ` `{` ` ` `$dp` `[` `$i` `][` `$j` `] = PHP_INT_MAX;` ` ` `// getting current string and ` ` ` `// current cost according to j` ` ` `if` `(` `$j` `== 0) ` ` ` `$curStr` `= ` `$arr` `[` `$i` `];` ` ` `else` ` ` `$curStr` `= ` `$revStr` `[` `$i` `];` ` ` ` ` `if` `(` `$j` `== 0) ` ` ` `$curCost` `= 0 ;` ` ` `else` ` ` `$curCost` `= ` `$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] + ` ` ` `$curCost` `);` ` ` `}` ` ` `}` ` ` `// getting minimum from both entries` ` ` `// of last index` ` ` `$res` `= min(` `$dp` `[` `$N` `- 1][0], ` `$dp` `[` `$N` `- 1][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` `?>` |

## Javascript

`<script>` `// Javascript 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.` ` ` `function` `minCost(arr,cost,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.` ` ` ` ` `let dp=` `new` `Array(N);` ` ` `for` `(let i=0;i<N;i++)` ` ` `{` ` ` `dp[i]=` `new` `Array(2);` ` ` `for` `(let j=0;j<2;j++)` ` ` `{` ` ` `dp[i][j]=0;` ` ` `}` ` ` `}` ` ` ` ` `// initializing dp array for first string` ` ` `dp[0][0] = 0;` ` ` `dp[0][1] = cost[0];` ` ` ` ` `// getting array of reversed strings` ` ` `let revStr = ` `new` `Array(N);` ` ` `for` `(let i=0;i<N;i++)` ` ` `{` ` ` `revStr[i]=` `""` `;` ` ` `}` ` ` ` ` `for` `(let i = 0; i < N; i++)` ` ` `{` ` ` `revStr[i] = arr[i];` ` ` `revStr[i] = reverse(revStr[i], 0, ` ` ` `revStr[i].length - 1);` ` ` `}` ` ` ` ` `let curStr = ` `""` `;` ` ` `let curCost;` ` ` ` ` `// looping for all strings` ` ` `for` `(let i = 1; i < N; i++)` ` ` `{` ` ` `// Looping twice, once for string and once` ` ` `// for reversed string` ` ` `for` `(let j = 0; j < 2; j++)` ` ` `{` ` ` `dp[i][j] = Number.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>=arr[i - 1]) ` ` ` `dp[i][j] = Math.min(dp[i][j], ` ` ` `dp[i - 1][0] + curCost);` ` ` `if` `(curStr>=revStr[i - 1]) ` ` ` `dp[i][j] = Math.min(dp[i][j],` ` ` `dp[i - 1][1] + curCost);` ` ` `}` ` ` `}` ` ` ` ` `// getting minimum from both entries of last index` ` ` `let res = Math.min(dp[N - 1][0], dp[N - 1][1]);` ` ` ` ` `return` `(res == Number.MAX_VALUE)? -1 : res;` ` ` `}` ` ` ` ` ` ` `function` `reverse(s,start,end)` ` ` `{` ` ` `// Temporary variable to store character ` ` ` `let temp;` ` ` `let str=s.split(` `""` `);` ` ` `while` `(start <= end) ` ` ` `{` ` ` ` ` `// Swapping the first and last character ` ` ` `temp = str[start];` ` ` `str[start] = str[end];` ` ` `str[end] = temp;` ` ` `start++;` ` ` `end--;` ` ` `}` ` ` `return` `str.toString();` ` ` ` ` `}` ` ` ` ` `// Driver Code` ` ` `let arr=[` `"aa"` `, ` `"ba"` `, ` `"ac"` `];` ` ` `let cost=[1, 3, 1];` ` ` `let N = arr.length;` ` ` `let res = minCost(arr, cost, N);` ` ` `if` `(res == -1)` ` ` `document.write(` `"Sorting not possible\n"` `);` ` ` `else` ` ` `document.write(` `"Minimum cost to "` `+ ` ` ` `"sort strings is "` `+ res);` ` ` ` ` ` ` `// This code is contributed by avanitrachhadiya2155` ` ` `</script>` |

**Output**

Minimum cost to sort strings is 1

**Time Complexity: O(N)****Auxiliary Space: O(N)**

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