Given two strings S1, S2 of length N and M respectively, and two positive integers N1 and N2, the task is to find the maximum count of non-overlapping subsequences of S1 which are equal to S2 by concatenating the string s1, n1 times and the string s2, n2 times.
Examples:
Input: S1 = “acb”, S2 = “ab”, N1 = 4, N2 = 2
Output: 2
Explanation:
Concatenating the string S1, N1 ( = 4) times modifies S1 to “acbacbacbacb”.
Concatenating the string S2, N2 ( = 2) times modifies S2 to “abab”.
Since the string S2 occurs twice as a non-overlapping subsequence in S1, the required output is 2.Input: S1 = “abc”, S2 = “a”, N1 = 1, N2 = 1
Output: 1
Approach: The problem can be solved using the concept of checking if a string is a subsequence of another string or not.
Follow the steps below to solve the problem:
- Iterate over the characters of the string S2 and check if all the characters of S2 are present in the string S1 or not. If found to be false, then no such subsequence of S1 is possible which can be made equal to S2 by concatenating the string S1, N1 times and the string S2, N2 times.
- Iterate over the characters of the string S1, N1 times circularly. For every ith index, check if any subsequence of S1 exists up to ith index, which is equal to S2 or not. If found to be true, then increment the count.
- Finally, print the count obtained divided by N2 as the required answer, after performing the above operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count maximum number of // occurrences of s2 as subsequence in s1 // by concatenating s1, n1 times and s2, n2 times int getMaxRepetitions(string s1, int n1, string s2, int n2) { int temp1[26] = {0}, temp2[26] = {0}; for ( char i:s1) temp1[i - 'a' ]++; for ( char i:s2) temp2[i - 'a' ]++; for ( int i = 0; i < 26; i++) { if (temp2[i] > temp1[i]) return 0; } // Stores number of times // s1 is traversed int s1_reps = 0; // Stores number of times // s2 is traversed int s2_reps = 0; // Mapping index of s2 to number // of times s1 and s2 are traversed map< int , pair< int , int > > s2_index_to_reps; s2_index_to_reps[0] = {0, 0}; // Stores index of s1 circularly int i = 0; // Stores index of s2 circularly int j = 0; // Traverse the string s1, n1 times while (s1_reps < n1){ // If current character of both // the string are equal if (s1[i] == s2[j]) // Update j j += 1; // Update i i += 1; // If j is length of s2 if (j == s2.size()) { // Update j for // circular traversal j = 0; // Update s2_reps s2_reps += 1; } // If i is length of s1 if (i == s1.size()) { // Update i for // circular traversal i = 0; // Update s1_reps s1_reps += 1; // If already mapped j // to (s1_reps, s2_reps) if (s2_index_to_reps.find(j) != s2_index_to_reps.end()) break ; // Mapping j to (s1_reps, s2_reps) s2_index_to_reps[j] = {s1_reps, s2_reps}; } } // If s1 already traversed n1 times if (s1_reps == n1) return s2_reps / n2; // Otherwise, traverse string s1 by multiple of // s1_reps and update both s1_reps and s2_reps int initial_s1_reps = s2_index_to_reps[j].first , initial_s2_reps = s2_index_to_reps[j].second; int loop_s1_reps = s1_reps - initial_s1_reps; int loop_s2_reps = s2_reps - initial_s2_reps; int loops = (n1 - initial_s1_reps); // Update s2_reps s2_reps = initial_s2_reps + loops * loop_s2_reps; // Update s1_reps s1_reps = initial_s1_reps + loops * loop_s1_reps; // If s1 is traversed less than n1 times while (s1_reps < n1) { // If current character in both // the string are equal if (s1[i] == s2[j]) // Update j j += 1; // Update i i += 1; // If i is length of s1 if (i == s1.size()) { // Update i for circular traversal i = 0; // Update s1_reps s1_reps += 1; } // If j is length of ss if (j == s2.size()) { // Update j for circular traversal j = 0; // Update s2_reps s2_reps += 1; } } return s2_reps / n2; } // Driver Code int main() { string s1 = "acb" ; int n1 = 4; string s2 = "ab" ; int n2 = 2; cout << (getMaxRepetitions(s1, n1, s2, n2)); return 0; } // This code is contributed by mohit kumar 29. |
Java
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class GFG{ // Function to count maximum number of // occurrences of s2 as subsequence in s1 // by concatenating s1, n1 times and s2, n2 times static int getMaxRepetitions(String s1, int n1, String s2, int n2) { int temp1[] = new int [ 26 ], temp2[] = new int [ 26 ]; for ( char i : s1.toCharArray()) temp1[i - 'a' ]++; for ( char i : s2.toCharArray()) temp2[i - 'a' ]++; for ( int i = 0 ; i < 26 ; i++) { if (temp2[i] > temp1[i]) return 0 ; } // Stores number of times // s1 is traversed int s1_reps = 0 ; // Stores number of times // s2 is traversed int s2_reps = 0 ; // Mapping index of s2 to number // of times s1 and s2 are traversed HashMap<Integer, int []> s2_index_to_reps = new HashMap<>(); s2_index_to_reps.put( 0 , new int [] { 0 , 0 }); // Stores index of s1 circularly int i = 0 ; // Stores index of s2 circularly int j = 0 ; // Traverse the string s1, n1 times while (s1_reps < n1) { // If current character of both // the string are equal if (s1.charAt(i) == s2.charAt(j)) // Update j j += 1 ; // Update i i += 1 ; // If j is length of s2 if (j == s2.length()) { // Update j for // circular traversal j = 0 ; // Update s2_reps s2_reps += 1 ; } // If i is length of s1 if (i == s1.length()) { // Update i for // circular traversal i = 0 ; // Update s1_reps s1_reps += 1 ; // If already mapped j // to (s1_reps, s2_reps) if (!s2_index_to_reps.containsKey(j)) break ; // Mapping j to (s1_reps, s2_reps) s2_index_to_reps.put( j, new int [] { s1_reps, s2_reps }); } } // If s1 already traversed n1 times if (s1_reps == n1) return s2_reps / n2; // Otherwise, traverse string s1 by multiple of // s1_reps and update both s1_reps and s2_reps int initial_s1_reps = s2_index_to_reps.get(j)[ 0 ], initial_s2_reps = s2_index_to_reps.get(j)[ 1 ]; int loop_s1_reps = s1_reps - initial_s1_reps; int loop_s2_reps = s2_reps - initial_s2_reps; int loops = (n1 - initial_s1_reps); // Update s2_reps s2_reps = initial_s2_reps + loops * loop_s2_reps; // Update s1_reps s1_reps = initial_s1_reps + loops * loop_s1_reps; // If s1 is traversed less than n1 times while (s1_reps < n1) { // If current character in both // the string are equal if (s1.charAt(i) == s2.charAt(j)) // Update j j += 1 ; // Update i i += 1 ; // If i is length of s1 if (i == s1.length()) { // Update i for circular traversal i = 0 ; // Update s1_reps s1_reps += 1 ; } // If j is length of ss if (j == s2.length()) { // Update j for circular traversal j = 0 ; // Update s2_reps s2_reps += 1 ; } } return s2_reps / n2; } // Driver Code public static void main(String[] args) { String s1 = "acb" ; int n1 = 4 ; String s2 = "ab" ; int n2 = 2 ; System.out.println( getMaxRepetitions(s1, n1, s2, n2)); } } // This code is contributed by Kingash |
Python3
# Python3 program for the above approach # Function to count maximum number of # occurrences of s2 as subsequence in s1 # by concatenating s1, n1 times and s2, n2 times def getMaxRepetitions(s1, n1, s2, n2): # If all the characters of s2 are not present in s1 if any (c for c in set (s2) if c not in set (s1)): return 0 # Stores number of times # s1 is traversed s1_reps = 0 # Stores number of times # s2 is traversed s2_reps = 0 # Mapping index of s2 to number # of times s1 and s2 are traversed s2_index_to_reps = { 0 : ( 0 , 0 )} # Stores index of s1 circularly i = 0 # Stores index of s2 circularly j = 0 # Traverse the string s1, n1 times while s1_reps < n1: # If current character of both # the string are equal if s1[i] = = s2[j]: # Update j j + = 1 # Update i i + = 1 # If j is length of s2 if j = = len (s2): # Update j for # circular traversal j = 0 # Update s2_reps s2_reps + = 1 # If i is length of s1 if i = = len (s1): # Update i for # circular traversal i = 0 # Update s1_reps s1_reps + = 1 # If already mapped j # to (s1_reps, s2_reps) if j in s2_index_to_reps: break # Mapping j to (s1_reps, s2_reps) s2_index_to_reps[j] = (s1_reps, s2_reps) # If s1 already traversed n1 times if s1_reps = = n1: return s2_reps / / n2 # Otherwise, traverse string s1 by multiple of # s1_reps and update both s1_reps and s2_reps initial_s1_reps, initial_s2_reps = s2_index_to_reps[j] loop_s1_reps = s1_reps - initial_s1_reps loop_s2_reps = s2_reps - initial_s2_reps loops = (n1 - initial_s1_reps) # Update s2_reps s2_reps = initial_s2_reps + loops * loop_s2_reps # Update s1_reps s1_reps = initial_s1_reps + loops * loop_s1_reps # If s1 is traversed less than n1 times while s1_reps < n1: # If current character in both # the string are equal if s1[i] = = s2[j]: # Update j j + = 1 # Update i i + = 1 # If i is length of s1 if i = = len (s1): # Update i for circular traversal i = 0 # Update s1_reps s1_reps + = 1 # If j is length of ss if j = = len (s2): # Update j for circular traversal j = 0 # Update s2_reps s2_reps + = 1 return s2_reps / / n2 # Driver Code if __name__ = = '__main__' : s1 = "acb" n1 = 4 s2 = "ab" n2 = 2 print (getMaxRepetitions(s1, n1, s2, n2)) |
Javascript
<script> // Javascript program for the above approach // Function to count maximum number of // occurrences of s2 as subsequence in s1 // by concatenating s1, n1 times and s2, n2 times function getMaxRepetitions(s1,n1,s2,n2) { let temp1 = new Array(26); let temp2 = new Array(26); for (let i = 0; i < 26; i++) { temp1[i] = 0; temp2[i] = 0; } for (let i = 0; i < s1.split( "" ).length; i++) temp1[s1[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; for (let i = 0; i < s2.split( "" ).length; i++) temp2[s2[i].charCodeAt(0) - 'a' .charCodeAt(0)]++; for (let i = 0; i < 26; i++) { if (temp2[i] > temp1[i]) return 0; } // Stores number of times // s1 is traversed let s1_reps = 0; // Stores number of times // s2 is traversed let s2_reps = 0; // Mapping index of s2 to number // of times s1 and s2 are traversed let s2_index_to_reps = new Map(); s2_index_to_reps.set(0, [ 0, 0 ]); // Stores index of s1 circularly let i = 0; // Stores index of s2 circularly let j = 0; // Traverse the string s1, n1 times while (s1_reps < n1) { // If current character of both // the string are equal if (s1[i] == s2[j]) // Update j j += 1; // Update i i += 1; // If j is length of s2 if (j == s2.length) { // Update j for // circular traversal j = 0; // Update s2_reps s2_reps += 1; } // If i is length of s1 if (i == s1.length) { // Update i for // circular traversal i = 0; // Update s1_reps s1_reps += 1; // If already mapped j // to (s1_reps, s2_reps) if (!s2_index_to_reps.has(j)) break ; // Mapping j to (s1_reps, s2_reps) s2_index_to_reps.set( j, [s1_reps, s2_reps ]); } } // If s1 already traversed n1 times if (s1_reps == n1) return s2_reps / n2; // Otherwise, traverse string s1 by multiple of // s1_reps and update both s1_reps and s2_reps let initial_s1_reps = s2_index_to_reps.get(j)[0], initial_s2_reps = s2_index_to_reps.get(j)[1]; let loop_s1_reps = s1_reps - initial_s1_reps; let loop_s2_reps = s2_reps - initial_s2_reps; let loops = (n1 - initial_s1_reps); // Update s2_reps s2_reps = initial_s2_reps + loops * loop_s2_reps; // Update s1_reps s1_reps = initial_s1_reps + loops * loop_s1_reps; // If s1 is traversed less than n1 times while (s1_reps < n1) { // If current character in both // the string are equal if (s1[i] == s2[j]) // Update j j += 1; // Update i i += 1; // If i is length of s1 if (i == s1.length) { // Update i for circular traversal i = 0; // Update s1_reps s1_reps += 1; } // If j is length of ss if (j == s2.length) { // Update j for circular traversal j = 0; // Update s2_reps s2_reps += 1; } } return s2_reps / n2; } // Driver Code let s1 = "acb" ; let n1 = 4; let s2 = "ab" ; let n2 = 2; document.write(getMaxRepetitions(s1, n1, s2, n2)); // This code is contributed by unknown2108 </script> |
C#
using System; using System.Collections.Generic; class GFG { // Function to count maximum number of // occurrences of s2 as subsequence in s1 // by concatenating s1, n1 times and s2, n2 times static int GetMaxRepetitions( string s1, int n1, string s2, int n2) { int [] temp1 = new int [26], temp2 = new int [26]; foreach ( char x in s1) temp1[x - 'a' ]++; foreach ( char x in s2) temp2[x - 'a' ]++; for ( int k = 0; k < 26; k++) { if (temp2[k] > temp1[k]) return 0; } // Stores number of times // s1 is traversed int s1Reps = 0; // Stores number of times // s2 is traversed int s2Reps = 0; // Mapping index of s2 to number // of times s1 and s2 are traversed Dictionary< int , int []> s2IndexToReps = new Dictionary< int , int []>(); s2IndexToReps[0] = new int [] { 0, 0 }; // Stores index of s1 circularly int i = 0; // Stores index of s2 circularly int j = 0; // Traverse the string s1, n1 times while (s1Reps < n1) { // If current character of both // the string are equal if (s1[i] == s2[j]) // Update j j += 1; // Update i i += 1; // If j is length of s2 if (j == s2.Length) { // Update j for // circular traversal j = 0; // Update s2Reps s2Reps += 1; } // If i is length of s1 if (i == s1.Length) { // Update i for // circular traversal i = 0; // Update s1Reps s1Reps += 1; // If already mapped j // to (s1Reps, s2Reps) if (!s2IndexToReps.ContainsKey(j)) break ; // Mapping j to (s1Reps, s2Reps) s2IndexToReps[j] = new int [] { s1Reps, s2Reps }; } } // If s1 already traversed n1 times if (s1Reps == n1) return s2Reps / n2; // Otherwise, traverse string s1 by multiple of // s1Reps and update both s1Reps and s2Reps int initialS1Reps = s2IndexToReps[j][0], initialS2Reps = s2IndexToReps[j][1]; int loop_s1Reps = s1Reps - initialS1Reps; int loop_s2Reps = s2Reps - initialS2Reps; int loops = (n1 - initialS1Reps); // Update s2Reps s2Reps = initialS2Reps + loops * loop_s2Reps; // Update s1Reps s1Reps = initialS1Reps + loops * loop_s1Reps; // If s1 is traversed less than n1 times while (s1Reps < n1) { // If current character in both // the string are equal if (s1[j] == s2[j]) // Update j j += 1; // Update i i += 1; // If i is length of s1 if (i == s1.Length) { // Update i for circular traversal i = 0; // Update s1Reps s1Reps += 1; } // If j is length of ss if (j == s2.Length) { // Update j for circular traversal j = 0; // Update s2Reps s2Reps += 1; } } return s2Reps / n2; } // Driver Code public static void Main( string [] args) { string s1 = "acb" ; int n1 = 4; string s2 = "ab" ; int n2 = 2; Console.WriteLine( GetMaxRepetitions(s1, n1, s2, n2)); } } |
2
Time Complexity: O((N + M) * n1)
Auxiliary Space:O(1)
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