Given two numeric strings, A and B. A numeric string is a string that contains only digits [‘0’-‘9’].
The task is to make both the strings equal in minimum cost. The only operation that you are allowed to do is to delete a character (i.e. digit) from any of the strings (A or B). The cost of deleting a digit D is D units.
Examples:
Input : A = “7135”, B = “135”
Output : 7
To make both string identical we have to delete ‘7’ from string A.Input : A = “9142”, B = “1429”
Output : 14
There are 2 ways to make string “9142” identical to “1429” i.e either by deleting ‘9’ from both the strings or by deleting ‘1’, ‘4’and ‘2’ from both the string. Deleting 142 from both the string will cost 2*(1+4+2)=14 which is more optimal than deleting ‘9’.
This problem is a variation of a popular Dynamic Programming problem – Longest Common Subsequence. The idea is to find the maximum weight common subsequence which will be our required optimal identical string. To find the cost of deletion, subtract the sum of maximum weight common subsequence from the sum of string A and B.
Minimum weight to make string identical = costA + costB – 2*(cost of LCS)
Implementation:
C++
// CPP program to find minimum cost to make// two numeric strings identical#include <bits/stdc++.h>using namespace std;typedef long long int ll;// Function to find weight of LCSint lcs(int dp[101][101], string a, string b, int m, int n){ for (int i = 0; i < 100; i++) for (int j = 0; j < 100; j++) dp[i][j] = -1; if (m < 0 || n < 0) { return 0; } // if this state is already // calculated then return if (dp[m][n] != -1) return dp[m][n]; int ans = 0; if (a[m] == b[n]) { // adding required weight for // particular match ans = int(a[m] - 48) + lcs(dp, a, b, m - 1, n - 1); } else // recurse for left and right child // and store the max ans = max(lcs(dp, a, b, m - 1, n), lcs(dp, a, b, m, n - 1)); dp[m][n] = ans; return ans;}// Function to calculate cost of stringint costOfString(string str){ int cost = 0; for (int i = 0; i < str.length(); i++) cost += int(str[i] - 48); return cost;}// Driver codeint main(){ string a, b; a = "9142"; b = "1429"; int dp[101][101]; // Minimum cost needed to make two strings identical cout << (costOfString(a) + costOfString(b) - 2 * lcs(dp, a, b, a.length() - 1, b.length() - 1)); return 0;} |
Java
// Java program to find minimum cost to make// two numeric strings identicalimport java.io.*;class GFG { // Function to find weight of LCS static int lcs(int dp[][], String a, String b, int m, int n){ for (int i = 0; i < 100; i++) for (int j = 0; j < 100; j++) dp[i][j] = -1; if (m < 0 || n < 0) { return 0; } // if this state is already // calculated then return if (dp[m][n] != -1) return dp[m][n]; int ans = 0; if (a.charAt(m) == b.charAt(n)) { // adding required weight for // particular match ans = (a.charAt(m) - 48) + lcs(dp, a, b, m - 1, n - 1); } else // recurse for left and right child // and store the max ans = Math.max(lcs(dp, a, b, m - 1, n), lcs(dp, a, b, m, n - 1)); dp[m][n] = ans; return ans;}// Function to calculate cost of string static int costOfString(String str){ int cost = 0; for (int i = 0; i < str.length(); i++) cost += (str.charAt(i) - 48); return cost;}// Driver code public static void main (String[] args) { String a, b; a = "9142"; b = "1429"; int dp[][] = new int[101][101]; // Minimum cost needed to make two strings identical System.out.print( (costOfString(a) + costOfString(b) - 2 * lcs(dp, a, b, a.length() - 1, b.length() - 1))); }}// This code is contributed by anuj_67. |
Python 3
# Python 3 program to find minimum cost # to make two numeric strings identical# Function to find weight of LCSdef lcs(dp, a, b, m, n): for i in range(100): for j in range(100): dp[i][j] = -1 if (m < 0 or n < 0) : return 0 # if this state is already calculated # then return if (dp[m][n] != -1): return dp[m][n] ans = 0 if (a[m] == b[n]): # adding required weight for # particular match ans = (ord(a[m]) - 48) + lcs(dp, a, b, m - 1, n - 1) else: # recurse for left and right child # and store the max ans = max(lcs(dp, a, b, m - 1, n), lcs(dp, a, b, m, n - 1)) dp[m][n] = ans return ans# Function to calculate cost of stringdef costOfString(s): cost = 0 for i in range(len(s)): cost += (ord(s[i]) - 48) return cost# Driver codeif __name__ == "__main__": a = "9142" b = "1429" dp = [[0 for x in range(101)] for y in range(101)] # Minimum cost needed to make two # strings identical print(costOfString(a) + costOfString(b) - 2 * lcs(dp, a, b, len(a) - 1, len(b) - 1))# This code is contributed by ita_c |
C#
// C# program to find minimum cost to make // two numeric strings identical using System;public class GFG { // Function to find weight of LCS static int lcs(int [,]dp, String a, String b, int m, int n) { for (int i = 0; i < 100; i++) for (int j = 0; j < 100; j++) dp[i,j] = -1; if (m < 0 || n < 0) { return 0; } // if this state is already // calculated then return if (dp[m,n] != -1) return dp[m,n]; int ans = 0; if (a[m] == b[n]) { // adding required weight for // particular match ans = (a[m] - 48) + lcs(dp, a, b, m - 1, n - 1); } else // recurse for left and right child // and store the max ans = Math.Max(lcs(dp, a, b, m - 1, n), lcs(dp, a, b, m, n - 1)); dp[m,n] = ans; return ans; } // Function to calculate cost of string static int costOfString(String str) { int cost = 0; for (int i = 0; i < str.Length; i++) cost += (str[i] - 48); return cost; } // Driver code public static void Main () { String a, b; a = "9142"; b = "1429"; int [,]dp = new int[101,101]; // Minimum cost needed to make two strings identical Console.Write( (costOfString(a) + costOfString(b) - 2 * lcs(dp, a, b, a.Length- 1, b.Length - 1))); } } // This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to find minimum cost to make// two numeric strings identical // Function to find weight of LCS function lcs(dp,a,b,m,n) { if (m < 0 || n < 0) { return 0; } // if this state is already // calculated then return if (dp[m][n] != -1) return dp[m][n]; let ans = 0; if (a[m] == b[n]) { // adding required weight for // particular match ans = (a[m].charCodeAt(0) - 48) + lcs(dp, a, b, m - 1, n - 1); } else // recurse for left and right child // and store the max ans = Math.max(lcs(dp, a, b, m - 1, n), lcs(dp, a, b, m, n - 1)); dp[m][n] = ans; return ans; } // Function to calculate cost of string function costOfString(str) { let cost = 0; for (let i = 0; i < str.length; i++) cost += (str[i].charCodeAt(0) - 48); return cost; } // Driver code let a = "9142"; let b = "1429"; let dp=new Array(101); for(let i=0;i<dp.length;i++) { dp[i]=new Array(101); for(let j=0;j<101;j++) { dp[i][j]=-1; } } // Minimum cost needed to make two strings identical document.write( (costOfString(a) + costOfString(b) - 2 * lcs(dp, a, b, a.length - 1, b.length - 1))); // This code is contributed by avanitrachhadiya2155</script> |
Output
14
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