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