Given two strings A and B and two integers b and m. The task is to find that if it is possible to form string B from A such that A is divided into groups of b characters except the last group which will have characters ? b and you are allowed to pick atmost m characters from each group, and also order of characters in B must be same as that of A. If it is possible then print Yes else print No.
Examples:
Input: A = abcbbcdefxyz, B = acdxyz, b = 5, m = 2
Output: Yes
Groups can be “abcbb”, “cdefx” and “yz”
Now “acdxyz” can be used to pick “ac” and “dx” can be picked from “cdefx”.
Finally, “yz” if the last group.Input: A = abcbbcdefxyz, B = baz, b = 3, m = 2
Output: No
Approach: The idea is to use binary search. Iterate through string A and store the frequency of each of the characters of A in vector S. Now iterate through B and if the current character is not in the vector then print No since its not possible to form string B using A. Else, check the first occurrence of current character starting from the index of the last chosen character low, which denotes starting position in string A from where we want to match characters of string B. Keep track of number of characters stored in each group. If it exceeds, the given limit of characters in current block, we update the pointer low to the next block.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if it is possible // to form B from A satisfying the given conditions bool isPossible(string A, string B, int b, int m) { // Vector to store the frequency // of characters in A vector< int > S[26]; // Vector to store the count of characters // used from a particular group of characters vector< int > box(A.length(), 0); // Store the frequency of the characters for ( int i = 0; i < A.length(); i++) { S[A[i] - 'a' ].push_back(i); } int low = 0; for ( int i = 0; i < B.length(); i++) { auto it = lower_bound(S[B[i] - 'a' ].begin(), S[B[i] - 'a' ].end(), low); // If a character in B is not // present in A if (it == S[B[i] - 'a' ].end()) return false ; int count = (*it) / b; box[count] = box[count] + 1; // If count of characters used from // a particular group of characters // exceeds m if (box[count] >= m) { count++; // Update low to the starting index // of the next group low = (count)*b; } // If count of characters used from // a particular group of characters // has not exceeded m else low = (*it) + 1; } return true ; } // Driver code int main() { string A = "abcbbcdefxyz" ; string B = "acdxyz" ; int b = 5; int m = 2; if (isPossible(A, B, b, m)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java implementation of the approach import java.io.*; import java.util.*; class GFG{ // Function that returns true if it is // possible to form B from A satisfying // the given conditions static boolean isPossible(String A, String B, int b, int m) { // List to store the frequency // of characters in A List<List<Integer>> S = new ArrayList<List<Integer>>(); for ( int i = 0 ; i < 26 ; i++) S.add( new ArrayList<Integer>()); // Vector to store the count of characters // used from a particular group of characters int [] box = new int [A.length()]; // Store the frequency of the characters for ( int i = 0 ; i < A.length(); i++) { S.get(A.charAt(i) - 'a' ).add(i); } int low = 0 ; for ( int i = 0 ; i < B.length(); i++) { List<Integer> indexes = S.get( B.charAt(i) - 'a' ); int it = lower_bound(indexes, low); // If a character in B is not // present in A if (it == indexes.size()) return false ; int count = indexes.get(it) / b; box[count] = box[count] + 1 ; // If count of characters used from // a particular group of characters // exceeds m if (box[count] >= m) { count++; // Update low to the starting index // of the next group low = (count) * b; } // If count of characters used from // a particular group of characters // has not exceeded m else low = indexes.get(it) + 1 ; } return true ; } static int lower_bound(List<Integer> indexes, int k) { int low = 0 , high = indexes.size() - 1 ; while (low < high) { int mid = (low + high) / 2 ; if (indexes.get(mid) < k) low = mid + 1 ; else high = mid; } return (indexes.get(low) < k) ? low + 1 : low; } // Driver code public static void main(String[] args) { String A = "abcbbcdefxyz" ; String B = "acdxyz" ; int b = 5 ; int m = 2 ; if (isPossible(A, B, b, m)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by jithin |
Python3
# Python3 implementation of the approach # Function that returns true if it is # possible to form B from A satisfying # the given conditions def isPossible(A, B, b, m) : # List to store the frequency # of characters in A S = [] for i in range ( 26 ) : S.append([]) # Vector to store the count of characters # used from a particular group of characters box = [ 0 ] * len (A) # Store the frequency of the characters for i in range ( len (A)) : S[ ord (A[i]) - ord ( 'a' )].append(i) low = 0 for i in range ( len (B)) : indexes = S[ ord (B[i]) - ord ( 'a' )] it = lower_bound(indexes, low) # If a character in B is not # present in A if (it = = len (indexes)) : return False count = indexes[it] / / b box[count] = box[count] + 1 # If count of characters used from # a particular group of characters # exceeds m if (box[count] > = m) : count + = 1 # Update low to the starting index # of the next group low = (count) * b # If count of characters used from # a particular group of characters # has not exceeded m else : low = indexes[it] + 1 return True def lower_bound(indexes, k) : low, high = 0 , len (indexes) - 1 while (low < high) : mid = (low + high) / / 2 if (indexes[mid] < k) : low = mid + 1 else : high = mid if indexes[low] < k : return (low + 1 ) else : return low A = "abcbbcdefxyz" B = "acdxyz" b = 5 m = 2 if (isPossible(A, B, b, m)) : print ( "Yes" ) else : print ( "No" ) # This code is contributed by divyeshrabadiya07 |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { // Function that returns true if it is // possible to form B from A satisfying // the given conditions static bool isPossible( string A, string B, int b, int m) { // List to store the frequency // of characters in A List<List< int >> S = new List<List< int >>(); for ( int i = 0; i < 26; i++) { S.Add( new List< int >()); } // Vector to store the count of characters // used from a particular group of characters int [] box = new int [A.Length]; // Store the frequency of the characters for ( int i = 0; i < A.Length; i++) { S[A[i] - 'a' ].Add(i); } int low = 0; for ( int i = 0; i < B.Length; i++) { List< int > indexes = S[B[i] - 'a' ]; int it = lower_bound(indexes, low); // If a character in B is not // present in A if (it == indexes.Count) return false ; int count = indexes[it] / b; box[count] = box[count] + 1; // If count of characters used from // a particular group of characters // exceeds m if (box[count] >= m) { count++; // Update low to the starting index // of the next group low = (count) * b; } // If count of characters used from // a particular group of characters // has not exceeded m else low = indexes[it] + 1; } return true ; } static int lower_bound(List< int > indexes, int k) { int low = 0, high = indexes.Count - 1; while (low < high) { int mid = (low + high) / 2; if (indexes[mid] < k) low = mid + 1; else high = mid; } return (indexes[low] < k) ? low + 1 : low; } static void Main() { string A = "abcbbcdefxyz" ; string B = "acdxyz" ; int b = 5; int m = 2; if (isPossible(A, B, b, m)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by divyesh072019 |
Javascript
// Function that returns true if it is // possible to form B from A satisfying // the given conditions const isPossible = (A, B, b, m) => { const S = []; // List to store the frequency // of characters in A for (let i = 0; i < 26; i++) { S.push([]); } // Vector to store the count of characters // used from a particular group of characters const box = new Array(A.length).fill(0); // Store the frequency of the characters for (let i = 0; i < A.length; i++) { S[A.charCodeAt(i) - 'a' .charCodeAt(0)].push(i); } let low = 0; for (let i = 0; i < B.length; i++) { const indexes = S[B.charCodeAt(i) - 'a' .charCodeAt(0)]; let it = lower_bound(indexes, low); // If a character in B is not // present in A if (it === indexes.length) { return false ; } let count = Math.floor(indexes[it] / b); box[count] = box[count] + 1; // If count of characters used from // a particular group of characters // exceeds m if (box[count] >= m) { // Update low to the starting index // of the next group count++; low = count * b; } // If count of characters used from // a particular group of characters // has not exceeded m else { low = indexes[it] + 1; } } return true ; }; const lower_bound = (indexes, k) => { let low = 0; let high = indexes.length - 1; while (low < high) { const mid = Math.floor((low + high) / 2); if (indexes[mid] < k) { low = mid + 1; } else { high = mid; } } return indexes[low] < k ? low + 1 : low; }; const A = 'abcbbcdefxyz' ; const B = 'acdxyz' ; const b = 5; const m = 2; console.log(isPossible(A, B, b, m) ? 'Yes' : 'No' ); |
Yes
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