Given two numeric strings S and Target, and integers X, M. The task is to check if string S can be converted into string Target using the below operation:
- In one operation it is required to replace each digit in S with its product with X, i.e S[i] = (S[I]*X).
- The above operation has to be done exactly M times.
Examples:
Input: S = “1234”, Target = “2550525100”, X = 5, M = 2
Output: YES
Explanation:
Initially S=1234
After the 1st, level S becomes =5101520
After the 2nd level, S=2550525100
Thus, we are able to make Target string So print YES
Input: S = “53783”, Target = “2893653”, X = 2, M = 3
Output: NO
Approach: This is an implementation-based problem simply traverse from left to right and multiply X in every digit which is present in the original string. Perform this M times with the given string S and check if it is possible to convert original string S into Target string.
Below is the implementation of the above approach:
C++
// C++ program to check// If strings are Equal or not#include <bits/stdc++.h>using namespace std;void makestring(string S, string Target, int X, int M){ // Out of X and M if any one // having value zero then just compare // original string To Target string if (X == 0 || M == 0) { if (S == Target) { cout << "YES" << endl; } } else { vector<string> v1; for (int i = 0; i < S.length(); i++) { // character to integer int val = S[i] - '0'; // convert into string string a = to_string(val); v1.push_back(a); } while (M--) { vector<string> v; for (int i = 0; i < v1.size(); i++) { for (int j = 0; j < v1[i].size(); j++) { int temp = v1[i][j] - '0'; // multiply X into given integer temp = temp * X; v.push_back(to_string(temp)); } } v1 = v; } // Store all character from vector // of string into temp string string temp = ""; for (int i = 0; i < v1.size(); i++) { for (int j = 0; j < v1[i].size(); j++) { temp.push_back(v1[i][j]); } } // Compare both Target and temp if (temp == Target) { cout << "YES" << endl; } else { cout << "NO" << endl; } }}// Driver Codeint main(){ string S = "1234"; string Target = "2550525100"; int X = 5; int M = 2; makestring(S, Target, X, M); return 0;} |
Java
import java.util.*;import java.io.*;// Java program for the above approachclass GFG{ static void makestring(String S, String Target, int X, int M) { // Out of X and M if any one // having value zero then just compare // original String To Target String if (X == 0 || M == 0) { if (S == Target) { System.out.println("YES"); } } else { ArrayList<String> v1 = new ArrayList<String>(); for (int i = 0 ; i < S.length() ; i++) { // character to integer int val = (int)S.charAt(i) - (int)'0'; // convert into String String a = Integer.toString(val); v1.add(a); } while (M > 0) { ArrayList<String> v = new ArrayList<String>(); for (int i = 0 ; i < v1.size() ; i++) { for (int j = 0; j < v1.get(i).length() ; j++) { int temp = (int)v1.get(i).charAt(j) - (int)'0'; // multiply X into given integer temp = temp * X; v.add(Integer.toString(temp)); } } v1 = v; M -= 1; } // Store all character from vector // of String into temp String String temp = ""; for (int i = 0 ; i < v1.size() ; i++) { for (int j = 0 ; j < v1.get(i).length() ; j++) { temp+=v1.get(i).charAt(j); } } // Compare both Target and temp if (temp.equals(Target)) { System.out.println("YES"); } else { System.out.println("NO"); } } } // Driver Code public static void main(String args[]) { String S = "1234"; String Target = "2550525100"; int X = 5; int M = 2; makestring(S, Target, X, M); }}// This code is contributed by subhamgoyal2014. |
Python3
# Python program to check# If strings are Equal or notdef makestring(S, Target, X, M): # Out of X and M if any one # having value zero then just compare # original string To Target string if (X == 0 or M == 0): if (S == Target): print("YES") else: v1 = [] for i in range(len(S)): # character to integer val = ord(S[i]) - ord('0') # convert into string a = str(val) v1.append(a) while (M): v = [] for i in range(len(v1)): for j in range(len(v1[i])): temp = ord(v1[i][j]) - ord('0') # multiply X into given integer temp = temp * X v.append(str(temp)) M -= 1 v1 = v.copy() # Store all character from vector # of string into temp string temp = "" for i in range(len(v1)): for j in range(len(v1[i])): temp += v1[i][j] # Compare both Target and temp if (temp == Target): print("YES") else: print("NO")# Driver CodeS = "1234"Target = "2550525100"X = 5M = 2makestring(S, Target, X, M)# This code is contributed by shinjanpatra |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ static void makestring(String S, String Target, int X, int M) { // Out of X and M if any one // having value zero then just compare // original String To Target String if (X == 0 || M == 0) { if (S == Target) { Console.WriteLine("YES"); } } else { List<String> v1 = new List<String>(); for (int i = 0; i < S.Length; i++) { // character to integer int val = (int)S[i] - (int)'0'; // convert into String String a = val.ToString(); v1.Add(a); } while (M > 0) { List<String> v = new List<String>(); for (int i = 0; i < v1.Count; i++) { for (int j = 0; j < v1[i].Length; j++) { int _temp = (int)v1[i][j] - (int)'0'; // multiply X into given integer _temp = _temp * X; v.Add(_temp.ToString()); } } v1 = v; M -= 1; } // Store all character from vector // of String into temp String String temp = ""; for (int i = 0; i < v1.Count; i++) { for (int j = 0; j < v1[i].Length; j++) { temp += v1[i][j]; } } // Compare both Target and temp if (temp.Equals(Target)) { Console.WriteLine("YES"); } else { Console.WriteLine("NO"); } } } // Driver Code public static void Main() { String S = "1234"; String Target = "2550525100"; int X = 5; int M = 2; makestring(S, Target, X, M); }}// This code is contributed by Saurabh Jaiswal. |
Javascript
<script> // JavaScript program to check // If strings are Equal or not const makestring = (S, Target, X, M) => { // Out of X and M if any one // having value zero then just compare // original string To Target string if (X == 0 || M == 0) { if (S == Target) { document.write("YES<br/>"); } } else { let v1 = []; for (let i = 0; i < S.length; i++) { // character to integer let val = S.charCodeAt(i) - '0'.charCodeAt(0); // convert into string let a = (val).toString(); v1.push(a); } while (M--) { let v = []; for (let i = 0; i < v1.length; i++) { for (let j = 0; j < v1[i].length; j++) { let temp = v1[i].charCodeAt(j) - '0'.charCodeAt(0); // multiply X into given integer temp = temp * X; v.push(temp.toString()); } } v1 = [...v]; } // Store all character from vector // of string into temp string let temp = ""; for (let i = 0; i < v1.length; i++) { for (let j = 0; j < v1[i].length; j++) { temp += v1[i][j]; } } // Compare both Target and temp if (temp == Target) { document.write("YES<br/>"); } else { document.write("NO<br/>"); } } } // Driver Code let S = "1234"; let Target = "2550525100"; let X = 5; let M = 2; makestring(S, Target, X, M); // This code is contributed by rakeshsahni</script> |
Output
YES
Time Complexity: O(N* N* M)
Auxiliary Space: O(1)
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