Given two strings X and Y consisting of only digits ‘0’ to ‘9’. Find minimum cost required to make the given two strings identical. Only operation allowed is to delete characters from any of the string. The cost of operation of deleting the digit ‘d’ is d units.
Input: X = 3759, Y = 9350 Output: 23 Explanation For making both string identical, delete characters 3, 7, 5 from first string and delete characters 3, 5, 0 from second string. Total cost of operation is 3 + 7 + 5 + 3 + 5 + 0 = 23 Input: X = 3198, Y = 98 Output: 4
This problem is a variation of Longest Common Subsequence( LCS ) and this one. The idea is simple, instead of finding the length of longest common subsequence, find the maximum cost by adding identical characters from both the string.
Now to find the minimum cost, subtract the above result from total cost of both strings i.e.,
costX = Cost of removing all characters
from string 'X'
CostY = Cost of removing all characters
from string 'Y'
cost_Id = Cost of removing identical characters
from both strings
Minimum cost to make both string identical =
costX + costY - cost_Id
Below is the implementation of above approach:
C++
/* C++ code to find minimum cost to make two strings identical */#include <bits/stdc++.h>using namespace std;/* Function to return cost of removing the identical characters in LCS for X[0..m-1], Y[0..n-1] */int lcs(char* X, char* Y, int m, int n){ int L[m + 1][n + 1]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains cost of removing identical characters in LCS of X[0..i-1] and Y[0..j-1] */ for (int i = 0; i <= m; ++i) { for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i][j] = 0; // If both characters are same, add both // of them else if (X[i - 1] == Y[j - 1]) L[i][j] = L[i - 1][j - 1] + 2 * (X[i - 1] - '0'); // Otherwise find the maximum cost among them else L[i][j] = max(L[i - 1][j], L[i][j - 1]); } } return L[m][n];}// Returns cost of making X[] and Y[] identicalint findMinCost(char X[], char Y[]){ // Find LCS of X[] and Y[] int m = strlen(X), n = strlen(Y); // Initialize the cost variable int cost = 0; // Find cost of all characters in // both strings for (int i = 0; i < m; ++i) cost += X[i] - '0'; for (int i = 0; i < n; ++i) cost += Y[i] - '0'; return cost - lcs(X, Y, m, n);}/* Driver program to test above function */int main(){ char X[] = "3759"; char Y[] = "9350"; cout << "Minimum Cost to make two strings " << "identical is = " << findMinCost(X, Y); return 0;} |
Java
// Java code to find minimum cost to // make two strings identicalimport java.util.*;import java.lang.*;public class GfG{ /* Function to return cost of removing the identicalcharacters in LCS for X[0..m-1], Y[0..n-1] */static int lcs(char[] X, char[] Y, int m, int n){ int[][] L=new int[m + 1][n + 1]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains cost of removing identical characters in LCS of X[0..i-1] and Y[0..j-1] */ for (int i = 0; i <= m; ++i) { for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i][j] = 0; // If both characters are same, // add both of them else if (X[i - 1] == Y[j - 1]) L[i][j] = L[i - 1][j - 1] + 2 * (X[i - 1] - '0'); // Otherwise find the maximum // cost among them else L[i][j] = L[i - 1][j] > L[i][j - 1] ? L[i - 1][j] : L[i][j - 1]; } } return L[m][n];}// Returns cost of making X[] and Y[] identicalstatic int findMinCost(char X[], char Y[]){ // Find LCS of X[] and Y[] int m = X.length, n = Y.length; // Initialize the cost variable int cost = 0; // Find cost of all characters in // both strings for (int i = 0; i < m; ++i) cost += X[i] - '0'; for (int i = 0; i < n; ++i) cost += Y[i] - '0'; return cost - lcs(X, Y, m, n);}// driver function public static void main(String argc[]){ char X[] = ("3759").toCharArray(); char Y[] = ("9350").toCharArray(); System.out.println("Minimum Cost to make two strings"+ " identical is = " +findMinCost(X, Y)); }}// This code is contributed by Prerna Saini |
Python3
# Python3 code to find minimum cost to make two strings # identical # Function to return cost of removing the identical # characters in LCS for X[0..m-1], Y[0..n-1] def lcs(X, Y, m, n): L=[[0 for i in range(n+1)]for i in range(m+1)] # Following steps build L[m+1][n+1] in bottom # up fashion. Note that L[i][j] contains cost # of removing identical characters in LCS of # X[0..i-1] and Y[0..j-1] for i in range(m+1): for j in range(n+1): if (i == 0 or j == 0): L[i][j] = 0 # If both characters are same, add both # of them elif (X[i - 1] == Y[j - 1]): L[i][j] = L[i - 1][j - 1] + 2 * (ord(X[i - 1]) - 48) # Otherwise find the maximum cost among them else: L[i][j] = max(L[i - 1][j], L[i][j - 1]) return L[m][n] # Returns cost of making X[] and Y[] identical def findMinCost( X, Y): # Find LCS of X[] and Y[] m = len(X) n = len(Y) # Initialize the cost variable cost = 0 # Find cost of all acters in # both strings for i in range(m): cost += ord(X[i]) - 48 for i in range(n): cost += ord(Y[i]) - 48 ans=cost - lcs(X, Y, m, n) return ans # Driver program to test above function X = "3759"Y = "9350"print("Minimum Cost to make two strings ", "identical is = " ,findMinCost(X, Y))#this code is contributed by sahilshelangia |
C#
// C# code to find minimum cost to // make two strings identicalusing System;public class GfG{ /* Function to return cost of removing the identical characters in LCS for X[0..m-1], Y[0..n-1] */ static int lcs(string X, string Y, int m, int n) { int [,]L=new int[m + 1,n + 1]; /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains cost of removing identical characters in LCS of X[0..i-1] and Y[0..j-1] */ for (int i = 0; i <= m; ++i) { for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i,j] = 0; // If both characters are same, // add both of them else if (X[i - 1] == Y[j - 1]) L[i,j] = L[i - 1,j - 1] + 2 * (X[i - 1] - '0'); // Otherwise find the maximum // cost among them else L[i,j] = L[i - 1,j] > L[i,j - 1] ? L[i - 1,j] : L[i,j - 1]; } } return L[m,n]; } // Returns cost of making X[] and Y[] identical static int findMinCost( string X, string Y) { // Find LCS of X[] and Y[] int m = X.Length, n = Y.Length; // Initialize the cost variable int cost = 0; // Find cost of all characters in // both strings for (int i = 0; i < m; ++i) cost += X[i] - '0'; for (int i = 0; i < n; ++i) cost += Y[i] - '0'; return cost - lcs(X, Y, m, n); } // Driver function public static void Main() { string X = "3759"; string Y= "9350"; Console.WriteLine("Minimum Cost to make two strings"+ " identical is = " +findMinCost(X, Y)); }}// This code is contributed by vt_m |
PHP
<?php// PHP code to find minimum cost to // make two strings identical/* Function to return cost of removing the identicalcharacters in LCS for X[0..m-1], Y[0..n-1] */function lcs( $X, $Y, $m, $n){ $L = array($m + 1,$n+ 1); /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains cost of removing identical characters in LCS of X[0..i-1] and Y[0..j-1] */ for ($i = 0; $i <= $m; ++$i) { for ($j = 0; $j <= $n; $j++) { if ($i == 0 || $j == 0) $L[$i][$j] = 0; // If both characters are same, // add both of them else if ($X[$i - 1] == $Y[$j - 1]) $L[$i][$j] = $L[$i - 1][$j - 1] + 2 * ($X[$i - 1] - '0'); // Otherwise find the maximum // cost among them else $L[$i][$j] = $L[$i - 1][$j] > $L[$i][$j - 1] ? $L[$i - 1][$j] : $L[$i][$j - 1]; } } return $L[$m][$n];}// Returns cost of making X[] and Y[] identicalfunction findMinCost($X, $Y){ // Find LCS of X[] and Y[] $m = sizeof($X); $n = sizeof($Y); // Initialize the cost variable $cost = 0; // Find cost of all characters in // both strings for ($i = 0; $i < $m; ++$i) $cost += $X[$i] - '0'; for ($i = 0; $i < $n; ++$i) $cost += $Y[$i] - '0'; return $cost - lcs($X, $Y, $m, $n);}// Driver code $X = str_split("3759"); $Y = str_split("9350"); echo("Minimum Cost to make two strings". " identical is = " .findMinCost($X, $Y)); // This code is contributed by Code_Mech. |
Javascript
<script> // Javascript code to find minimum cost to make two strings identical /* Function to return cost of removing the identical characters in LCS for X[0..m-1], Y[0..n-1] */ function lcs(X, Y, m, n) { let L=new Array(m + 1); for (let i = 0; i <= m; ++i) { L[i] = new Array(n + 1); for (let j = 0; j <= n; j++) { L[i][j] = 0; } } /* Following steps build L[m+1][n+1] in bottom up fashion. Note that L[i][j] contains cost of removing identical characters in LCS of X[0..i-1] and Y[0..j-1] */ for (let i = 0; i <= m; ++i) { for (let j = 0; j <= n; j++) { if (i == 0 || j == 0) L[i][j] = 0; // If both characters are same, // add both of them else if (X[i - 1] == Y[j - 1]) L[i][j] = L[i - 1][j - 1] + 2 * (X[i - 1] - '0'); // Otherwise find the maximum // cost among them else L[i][j] = L[i - 1][j] > L[i][j - 1] ? L[i - 1][j] : L[i][j - 1]; } } return L[m][n]; } // Returns cost of making X[] and Y[] identical function findMinCost(X, Y) { // Find LCS of X[] and Y[] let m = X.length, n = Y.length; // Initialize the cost variable let cost = 0; // Find cost of all characters in // both strings for (let i = 0; i < m; ++i) cost += X[i].charCodeAt() - '0'.charCodeAt(); for (let i = 0; i < n; ++i) cost += Y[i].charCodeAt() - '0'.charCodeAt(); return cost - lcs(X, Y, m, n); } let X = ("3759").split(''); let Y = ("9350").split(''); document.write("Minimum Cost to make two strings"+ " identical is = " +findMinCost(X, Y)); </script> |
Output
Minimum Cost to make two strings identical is = 23
Time complexity: O(m*n), As we are traversing the matrix of size m*n once.
Auxiliary space: O(m*n), Extra space is used to store the elements of the matrix.
Efficient approach: Space optimization
In previous approach the current value L[i][j] is only depend upon the current and previous row values of matrix. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector L of size n+1 and initialize it with 0.
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create variable prev used to store previous computations and curr to store the current value.
- After every iteration assign the value of curr to prev for further iteration.
- Initialize a variable cost to store the final answer and update it by iterating through the L.
- At last return and print the final answer stored in cost.
Implementation:
C++
// C++ code to find minimum cost to // make two strings identical#include <bits/stdc++.h>using namespace std;// Returns cost of making X[] and Y[] identicalint findMinCost(char X[], char Y[]) { int m = strlen(X), n = strlen(Y); // vector to store computations vector<int> L(n + 1, 0); // iterate over subproblems for (int i = 0; i < m; ++i) { // to store previous element int prev = 0; for (int j = 0; j < n; j++) { // current element int curr = L[j+1]; // update L if (X[i] == Y[j]) { L[j+1] = prev + 2 * (X[i] - '0'); } else { L[j+1] = max(L[j], L[j+1]); } // assign value to iterate further prev = curr; } } // to store minimum cost int cost = 0; for (int i = 0; i < m; ++i) { cost += X[i] - '0'; } for (int i = 0; i < n; ++i) { cost += Y[i] - '0'; } // return final answer return cost - L[n];}// Driver codeint main() { char X[] = "3759"; char Y[] = "9350"; // function call cout << "Minimum Cost to make two strings " << "identical is = " << findMinCost(X, Y); return 0;} |
Java
// Java code to find minimum cost to// make two strings identicalimport java.lang.*;import java.util.*;class GfG { // Returns cost of making X[] and Y[] identical static int findMinCost(char X[], char Y[]) { int m = X.length, n = Y.length; // vector to store computations int[] L = new int[n + 1]; // iterate over subproblems for (int i = 0; i < m; ++i) { // to store previous element int prev = 0; for (int j = 0; j < n; j++) { // current element int curr = L[j + 1]; // update L if (X[i] == Y[j]) { L[j + 1] = prev + 2 * (X[i] - '0'); } else { L[j + 1] = Math.max(L[j], L[j + 1]); } // assign value to iterate further prev = curr; } } // to store minimum cost int cost = 0; for (int i = 0; i < m; ++i) { cost += X[i] - '0'; } for (int i = 0; i < n; ++i) { cost += Y[i] - '0'; } // return final answer return cost - L[n]; } public static void main(String[] args) { char[] X = "3759".toCharArray(); char[] Y = "9350".toCharArray(); // function call System.out.println( "Minimum Cost to make two strings identical is = " + findMinCost(X, Y)); }}// This code is contributed by Karandeep1234 |
Python
def find_min_cost(X, Y): m, n = len(X), len(Y) # list to store computations L = [0] * (n + 1) # iterate over subproblems for i in range(m): # to store previous element prev = 0 for j in range(n): # current element curr = L[j+1] # update L if X[i] == Y[j]: L[j+1] = prev + 2 * int(X[i]) else: L[j+1] = max(L[j], L[j+1]) # assign value to iterate further prev = curr # to store minimum cost cost = sum(int(c) for c in X) + sum(int(c) for c in Y) # return final answer return cost - L[n]# Driver codeX = "3759"Y = "9350"# function callprint("Minimum Cost to make two strings identical is = " + str(find_min_cost(X, Y))) |
C#
using System;using System.Collections.Generic;class Program { // Returns cost of making X[] and Y[] identical static int findMinCost(char[] X, char[] Y) { int m = X.Length, n = Y.Length; // vector to store computations List<int> L = new List<int>(n + 1); for (int i = 0; i < n + 1; i++) { L.Add(0); } // iterate over subproblems for (int i = 0; i < m; ++i) { // to store previous element int prev = 0; for (int j = 0; j < n; j++) { // current element int curr = L[j + 1]; // update L if (X[i] == Y[j]) { L[j + 1] = prev + 2 * (X[i] - '0'); } else { L[j + 1] = Math.Max(L[j], L[j + 1]); } // assign value to iterate further prev = curr; } } // to store minimum cost int cost = 0; for (int i = 0; i < m; ++i) { cost += X[i] - '0'; } for (int i = 0; i < n; ++i) { cost += Y[i] - '0'; } // return final answer return cost - L[n]; } // Driver code static void Main(string[] args) { char[] X = "3759".ToCharArray(); char[] Y = "9350".ToCharArray(); // function call Console.WriteLine( "Minimum Cost to make two strings identical is = " + findMinCost(X, Y)); }} |
Javascript
function findMinCost(X, Y) { const m = X.length; const n = Y.length; // list to store computations const L = new Array(n + 1).fill(0); // iterate over subproblems for (let i = 0; i < m; i++) { // to store previous element let prev = 0; for (let j = 0; j < n; j++) { // current element const curr = L[j+1]; // update L if (X[i] === Y[j]) { L[j+1] = prev + 2 * parseInt(X[i]); } else { L[j+1] = Math.max(L[j], L[j+1]); } // assign value to iterate further prev = curr; } } // to store minimum cost let cost = X.split("").reduce((acc, c) => acc + parseInt(c), 0) + Y.split("").reduce((acc, c) => acc + parseInt(c), 0); // return final answer return cost - L[n];}// Driver codeconst X = "3759";const Y = "9350";// function callconsole.log("Minimum Cost to make two strings identical is = "+ findMinCost(X, Y)); |
Output
Minimum Cost to make two strings identical is = 23
Time complexity: O(m*n), As we are traversing the matrix of size m*n once.
Auxiliary space: O(n)
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