Given a string str of lowercase English alphabets and an integer m. The task is to count how many positions are there in the string such that if you partition the string into two non-empty sub-strings, there are at least m characters with the same frequency in both the sub-strings.
The characters need to be present in the string str.
Examples:
Input: str = “aabbccaa”, m = 2
Output: 2
The string has length 8, so there are 7 positions available to perform the partition.
i.e. a|a|b|b|c|c|a|a
Only two partitions are possible which satisfy the given constraints.
aab|bccaa – On the left half of the separator, ‘a’ has frequency 2 and ‘b’ has frequency 1
which is same as that of the right half.
aabbc|caa – On the left half of the separator, ‘a’ has frequency 2 and ‘c’ has frequency 1
which is same as that of the right half.
Input: str = “aabbaa”, m = 2
Output: 1
Approach: For each partition position, calculate the frequencies of each of the characters of the string in both the partitions. Then calculate the number of characters having same frequency in both partitions. If the count of such characters is at least m then add 1 to the required count of partitions.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the number of ways// to partition the given so that the// given condition is satisfiedint countWays(string str, int m){ // Hashset to store unique characters // in the given string set<char> s; for (int i = 0; i < str.length(); i++) s.insert(str[i]); // To store the number of ways // to partition the string int result = 0; for (int i = 1; i < str.length(); i++) { // Hashmaps to store frequency of characters // of both the partitions map<char, int> first_map, second_map; // Iterate in the first partition for (int j = 0; j < i; j++) // If character already exists in the hashmap // then increase it's frequency first_map[str[j]]++; // Iterate in the second partition for (int k = 0; k < str.length(); k++) // If character already exists in the hashmap // then increase it's frequency second_map[str[k]]++; // Iterator for HashSet set<char>::iterator itr = s.begin(); // To store the count of characters that have // equal frequencies in both the partitions int total_count = 0; while (++itr != s.end()) { // first_count and second_count keeps track // of the frequencies of each character int first_count = 0, second_count = 0; char ch = *(itr); // Frequency of the character // in the first partition if (first_map.find(ch) != first_map.end()) first_count = first_map[ch]; // Frequency of the character // in the second partition if (second_map.find(ch) != second_map.end()) second_count = second_map[ch]; // Check if frequency is same // in both the partitions if (first_count == second_count && first_count != 0) total_count += 1; } // Check if the condition is satisfied // for the current partition if (total_count >= m) result += 1; } return result;}// Driver codeint main(int argc, char const *argv[]){ string str = "aabbccaa"; int m = 2; cout << countWays(str, m) << endl; return 0;}// This code is contributed by// sanjeev2552 |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to return the number of ways // to partition the given so that the // given condition is satisfied static int countWays(String str, int m) { // Hashset to store unique characters // in the given string HashSet<Character> set = new HashSet<Character>(); for (int i = 0; i < str.length(); i++) set.add(str.charAt(i)); // To store the number of ways // to partition the string int result = 0; for (int i = 1; i < str.length(); i++) { // Hashmaps to store frequency of characters // of both the partitions HashMap<Character, Integer> first_map = new HashMap<Character, Integer>(); HashMap<Character, Integer> second_map = new HashMap<Character, Integer>(); // Iterate in the first partition for (int j = 0; j < i; j++) { // If character already exists in the hashmap // then increase it's frequency if (first_map.containsKey(str.charAt(j))) first_map.put(str.charAt(j), (first_map.get(str.charAt(j)) + 1)); // Else create an entry for it in the Hashmap else first_map.put(str.charAt(j), 1); } // Iterate in the second partition for (int k = i; k < str.length(); k++) { // If character already exists in the hashmap // then increase it's frequency if (second_map.containsKey(str.charAt(k))) second_map.put(str.charAt(k), (second_map.get(str.charAt(k)) + 1)); // Else create an entry for it in the Hashmap else second_map.put(str.charAt(k), 1); } // Iterator for HashSet Iterator itr = set.iterator(); // To store the count of characters that have // equal frequencies in both the partitions int total_count = 0; while (itr.hasNext()) { // first_count and second_count keeps track // of the frequencies of each character int first_count = 0, second_count = 0; char ch = (char)itr.next(); // Frequency of the character // in the first partition if (first_map.containsKey(ch)) first_count = first_map.get(ch); // Frequency of the character // in the second partition if (second_map.containsKey(ch)) second_count = second_map.get(ch); // Check if frequency is same in both the partitions if (first_count == second_count && first_count != 0) total_count += 1; } // Check if the condition is satisfied // for the current partition if (total_count >= m) result += 1; } return result; } // Driver code public static void main(String[] args) { String str = "aabbccaa"; int m = 2; System.out.println(countWays(str, m)); }} |
Python3
# Python3 implementation of the approach from collections import defaultdict# Function to return the number of ways # to partition the given so that the # given condition is satisfied def countWays(string, m): # Hashset to store unique # characters in the given string Set = set() for i in range(0, len(string)): Set.add(string[i]) # To store the number of ways # to partition the string result = 0 for i in range(1, len(string)): # Hashmaps to store frequency of # characters of both the partitions first_map = defaultdict(lambda:0) second_map = defaultdict(lambda:0) # Iterate in the first partition for j in range(0, i): first_map[string[j]] += 1 # Iterate in the second partition for k in range(i, len(string)): second_map[string[k]] += 1 # To store the count of characters that have # equal frequencies in both the partitions total_count = 0 for ch in Set: # first_count and second_count keeps track # of the frequencies of each character first_count, second_count = 0, 0 # Frequency of the character # in the first partition if ch in first_map: first_count = first_map[ch] # Frequency of the character # in the second partition if ch in second_map: second_count = second_map[ch] # Check if frequency is same in both the partitions if first_count == second_count and first_count != 0: total_count += 1 # Check if the condition is satisfied # for the current partition if total_count >= m: result += 1 return result # Driver code if __name__ == "__main__": string = "aabbccaa" m = 2 print(countWays(string, m)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;public class GFG { // Function to return the number of ways // to partition the given so that the // given condition is satisfied static int countWays(String str, int m) { // Hashset to store unique characters // in the given string HashSet<char> set = new HashSet<char>(); for (int i = 0; i < str.Length; i++) set.Add(str[i]); // To store the number of ways // to partition the string int result = 0; for (int i = 1; i < str.Length; i++) { // Hashmaps to store frequency of characters // of both the partitions Dictionary<char, int> first_map = new Dictionary<char, int>(); Dictionary<char, int> second_map = new Dictionary<char, int>(); // Iterate in the first partition for (int j = 0; j < i; j++) { // If character already exists in the hashmap // then increase it's frequency if (first_map.ContainsKey(str[j])) first_map[str[j]] = (first_map[str[j]] + 1); // Else create an entry for it in the Hashmap else first_map.Add(str[j], 1); } // Iterate in the second partition for (int k = i; k < str.Length; k++) { // If character already exists in the hashmap // then increase it's frequency if (second_map.ContainsKey(str[k])) second_map[str[k]] = (second_map[str[k]] + 1); // Else create an entry for it in the Hashmap else second_map.Add(str[k], 1); } // To store the count of characters that have // equal frequencies in both the partitions int total_count = 0; // Iterator for HashSet foreach (int itr in set) { // first_count and second_count keeps track // of the frequencies of each character int first_count = 0, second_count = 0; char ch = (char)itr; // Frequency of the character // in the first partition if (first_map.ContainsKey(ch)) first_count = first_map[ch]; // Frequency of the character // in the second partition if (second_map.ContainsKey(ch)) second_count = second_map[ch]; // Check if frequency is same in both the partitions if (first_count == second_count && first_count != 0) total_count += 1; } // Check if the condition is satisfied // for the current partition if (total_count >= m) result += 1; } return result; } // Driver code public static void Main(String[] args) { String str = "aabbccaa"; int m = 2; Console.WriteLine(countWays(str, m)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Function to return the number of ways// to partition the given so that the// given condition is satisfiedfunction countWays(str, m){ // Hashset to store unique characters // in the given string var s = new Set(); for (var i = 0; i < str.length; i++) s.add(str[i]); // To store the number of ways // to partition the string var result = 0; for (var i = 1; i < str.length; i++) { // Hashmaps to store frequency of characters // of both the partitions var first_map = new Map(), second_map = new Map(); // Iterate in the first partition for (var j = 0; j < i; j++) // If character already exists in the hashmap // then increase it's frequency if(first_map.has(str[j])) first_map.set(str[j], first_map.get(str[j])+1) else first_map.set(str[j], 1) // Iterate in the second partition for (var k = 0; k < str.length; k++) // If character already exists in the hashmap // then increase it's frequency if(second_map.has(str[k])) second_map.set(str[k], second_map.get(str[k])+1) else second_map.set(str[k], 1) // To store the count of characters that have // equal frequencies in both the partitions var total_count = 0; s.forEach(itr => { // first_count and second_count keeps track // of the frequencies of each character var first_count = 0, second_count = 0; var ch = itr; // Frequency of the character // in the first partition if (first_map.has(ch)) first_count = first_map.get(ch); // Frequency of the character // in the second partition if (second_map.has(ch)) second_count = second_map.get(ch); // Check if frequency is same // in both the partitions if (first_count == second_count && first_count != 0) total_count += 1; }); // Check if the condition is satisfied // for the current partition if (total_count >= m) result += 1; } return result;}// Driver codevar str = "aabbccaa";var m = 2;document.write( countWays(str, m));// This code is contributed by itsok.</script> |
Output:
2
Time Complexity: O(n*n*log(n)), as nested loops are used for iteration
Auxiliary Space: O(n), as extra space of size n is used to make a set and map
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
