Count of Substrings with at least K pairwise Distinct Characters having same Frequency
Last Updated :
06 May, 2021
Given a string S and an integer K, the task is to find the number of substrings which consists of at least K pairwise distinct characters having same frequency.
Examples:
Input: S = "abasa", K = 2
Output: 5
Explanation:
The substrings in having 2 pairwise distinct characters with same frequency are {"ab", "ba", "as", "sa", "bas"}.
Input: S = "abhay", K = 3
Output: 4
Explanation:
The substrings having 3 pairwise distinct characters with same frequency are {"abh", "bha", "hay", "bhay"}.
Naive Approach: The simplest approach to solve this problem is to generate all possible substrings of the given string and check if both the conditions are satisfied. If found to be true, increase count. Finally, print count.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the steps below to solve the problem:
- Check if the frequencies of each character is same. If found to be true, simply generate all the substrings to check if each character satisfies the condition of at least N pairwise distinct characters.
- Precompute the frequencies of characters to check the conditions for each substring.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the substring with K
// pairwise distinct characters and
// with same frequency
int no_of_substring(string s, int N)
{
// Stores the occurrence of each
// character in the substring
int fre[26];
int str_len;
// Length of the string
str_len = (int)s.length();
int count = 0;
// Iterate over the string
for (int i = 0; i < str_len; i++) {
// Set all values at each index to zero
memset(fre, 0, sizeof(fre));
int max_index = 0;
// Stores the count of
// unique characters
int dist = 0;
// Moving the substring ending at j
for (int j = i; j < str_len; j++) {
// Calculate the index of
// character in frequency array
int x = s[j] - 'a';
if (fre[x] == 0)
dist++;
// Increment the frequency
fre[x]++;
// Update the maximum index
max_index = max(max_index, fre[x]);
// Check for both the conditions
if (dist >= N && ((max_index * dist)
== (j - i + 1)))
count++;
}
}
// Return the answer
return count;
}
// Driver Code
int main()
{
string s = "abhay";
int N = 3;
// Function call
cout << no_of_substring(s, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the subString with K
// pairwise distinct characters and
// with same frequency
static int no_of_subString(String s, int N)
{
// Stores the occurrence of each
// character in the subString
int fre[] = new int[26];
int str_len;
// Length of the String
str_len = (int)s.length();
int count = 0;
// Iterate over the String
for(int i = 0; i < str_len; i++)
{
// Set all values at each index to zero
Arrays.fill(fre, 0);
int max_index = 0;
// Stores the count of
// unique characters
int dist = 0;
// Moving the subString ending at j
for(int j = i; j < str_len; j++)
{
// Calculate the index of
// character in frequency array
int x = s.charAt(j) - 'a';
if (fre[x] == 0)
dist++;
// Increment the frequency
fre[x]++;
// Update the maximum index
max_index = Math.max(max_index, fre[x]);
// Check for both the conditions
if (dist >= N && ((max_index * dist) ==
(j - i + 1)))
count++;
}
}
// Return the answer
return count;
}
// Driver Code
public static void main(String[] args)
{
String s = "abhay";
int N = 3;
// Function call
System.out.print(no_of_subString(s, N));
}
}
// This code is contributed by Amit Katiyar
# Python3 program to implement
# the above approach
# Function to find the substring with K
# pairwise distinct characters and
# with same frequency
def no_of_substring(s, N):
# Length of the string
str_len = len(s)
count = 0
# Iterate over the string
for i in range(str_len):
# Stores the occurrence of each
# character in the substring
# Set all values at each index to zero
fre = [0] * 26
max_index = 0
# Stores the count of
# unique characters
dist = 0
# Moving the substring ending at j
for j in range(i, str_len):
# Calculate the index of
# character in frequency array
x = ord(s[j]) - ord('a')
if (fre[x] == 0):
dist += 1
# Increment the frequency
fre[x] += 1
# Update the maximum index
max_index = max(max_index, fre[x])
# Check for both the conditions
if(dist >= N and
((max_index * dist) == (j - i + 1))):
count += 1
# Return the answer
return count
# Driver Code
s = "abhay"
N = 3
# Function call
print(no_of_substring(s, N))
# This code is contributed by Shivam Singh
// C# program for the above approach
using System;
class GFG{
// Function to find the subString with K
// pairwise distinct characters and
// with same frequency
static int no_of_subString(String s, int N)
{
// Stores the occurrence of each
// character in the subString
int []fre = new int[26];
int str_len;
// Length of the String
str_len = (int)s.Length;
int count = 0;
// Iterate over the String
for(int i = 0; i < str_len; i++)
{
// Set all values at each index to zero
fre = new int[26];
int max_index = 0;
// Stores the count of
// unique characters
int dist = 0;
// Moving the subString ending at j
for(int j = i; j < str_len; j++)
{
// Calculate the index of
// character in frequency array
int x = s[j] - 'a';
if (fre[x] == 0)
dist++;
// Increment the frequency
fre[x]++;
// Update the maximum index
max_index = Math.Max(max_index, fre[x]);
// Check for both the conditions
if (dist >= N && ((max_index * dist) ==
(j - i + 1)))
count++;
}
}
// Return the answer
return count;
}
// Driver Code
public static void Main(String[] args)
{
String s = "abhay";
int N = 3;
// Function call
Console.Write(no_of_subString(s, N));
}
}
// This code is contributed by Amit Katiyar
<script>
// javascript program for the above approach
// Function to find the subString with K
// pairwise distinct characters and
// with same frequency
function no_of_subString(s , N)
{
// Stores the occurrence of each
// character in the subString
var fre = Array.from({length: 26}, (_, i) => 0);
var str_len;
// Length of the String
str_len = parseInt(s.length);
var count = 0;
// Iterate over the String
for(i = 0; i < str_len; i++)
{
// Set all values at each index to zero
fre = Array(26).fill(0);
var max_index = 0;
// Stores the count of
// unique characters
var dist = 0;
// Moving the subString ending at j
for(j = i; j < str_len; j++)
{
// Calculate the index of
// character in frequency array
var x = s.charAt(j).charCodeAt(0) - 'a'.charCodeAt(0);
if (fre[x] == 0)
dist++;
// Increment the frequency
fre[x]++;
// Update the maximum index
max_index = Math.max(max_index, fre[x]);
// Check for both the conditions
if (dist >= N && ((max_index * dist) ==
(j - i + 1)))
count++;
}
}
// Return the answer
return count;
}
// Driver Code
var s = "abhay";
var N = 3;
// Function call
document.write(no_of_subString(s, N));
// This code contributed by shikhasingrajput
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
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