Minimum cost to sort strings using reversal operations of different costs

• Difficulty Level : Hard
• Last Updated : 14 Apr, 2021

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] 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++ 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 methodsint 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 operationimport 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];     // initializing dp array for first string    dp = 0;    dp = 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] = 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] + curCost);            if (curStr.compareTo(revStr[i - 1]) >= 0)                dp[i][j] = Math.min(dp[i][j],                                    dp[i - 1] + curCost);        }    }     // getting minimum from both entries of last index    int res = Math.min(dp[N - 1], dp[N - 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 Codepublic 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     dp = reverseCost     # 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] + curCost)             if (curStr >= rev_arr[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])     return res if res != float("Inf") else -1  # Driver codedef 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 operationusing 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 Codepublic 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?>

Javascript



Output:

Minimum cost to sort strings is 1

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