Skip to content
Related Articles

Related Articles

Improve Article

Count substrings having frequency of a character exceeding that of another character in a string

  • Difficulty Level : Hard
  • Last Updated : 26 Aug, 2021
Geek Week

Given a string S of size N consisting of characters a, b, and c only, the task is to find the number of substrings of the given string S such that the frequency of character a is greater than the frequency of character c.

Examples:

Input: S = “abcc”
Output: 2
Explanation:
Below are all the possible substrings of S(= “abcc”) having the frequency of the character greater than the character c:

  1. “a”: The frequency of a and c is 1 and 0 respectively.
  2. “ab”: The frequency of a and c is 1 and 0 respectively.

Therefore, the count of such substrings is 2.

Input: S = “abcabcabcaaaaabbbccccc”
Output: 148

Naive Approach: The simplest approach to solve the given problem is to generate all possible substrings of the given string S and count those substrings having a count of character ‘a’ greater than the count of character ‘c’. After checking for all the substrings, print the value of the total count as the result.

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 number of
// substrings having the frequency of
// 'a' greater than frequency of 'c'
void countSubstrings(string& s)
{
    // Stores the size of the string
    int n = s.length();
 
    // Stores the resultant
    // count of substrings
    int ans = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++) {
 
        // Store the difference between
        // frequency of 'a' and 'c'
        int cnt = 0;
 
        // Traverse all substrings
        // beginning at index i
        for (int j = i; j < n; j++) {
            if (s[j] == 'a')
                cnt++;
            else if (s[j] == 'c')
                cnt--;
 
            // If the frequency of 'a'
            // is greater than 'c'
            if (cnt > 0) {
                ans++;
            }
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Drive Code
int main()
{
    string S = "abccaab";
    countSubstrings(S);
 
    return 0;
}

Java




// Java program for the above approach
public class GFG
{
     
// Function to find the number of
// substrings having the frequency of
// 'a' greater than frequency of 'c'
public static void countSubstrings(String s)
{
   
    // Stores the size of the string
    int n = s.length();
 
    // Stores the resultant
    // count of substrings
    int ans = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++) {
 
        // Store the difference between
        // frequency of 'a' and 'c'
        int cnt = 0;
 
        // Traverse all substrings
        // beginning at index i
        for (int j = i; j < n; j++) {
            if (s.charAt(j) == 'a')
                cnt++;
            else if (s.charAt(j) == 'c')
                cnt--;
 
            // If the frequency of 'a'
            // is greater than 'c'
            if (cnt > 0) {
                ans++;
            }
        }
    }
 
    // Print the answer
    System.out.println(ans);
}
 
// Drive Code
public static void main(String args[])
{
    String S = "abccaab";
    countSubstrings(S);
 
}
}
 
// This code is contributed by SoumikMondal

Python3




# python program for the above approach
 
# Function to find the number of
# substrings having the frequency of
# 'a' greater than frequency of 'c'
def countSubstrings(s):
   
    # Stores the size of the string
    n = len(s)
 
    # Stores the resultant
    # count of substrings
    ans = 0
 
    # Traverse the given string
    for i in range(n):
       
        # Store the difference between
        # frequency of 'a' and 'c'
        cnt = 0
 
        # Traverse all substrings
        # beginning at index i
        for j in range(i, n):
            if (s[j] == 'a'):
                cnt += 1
            elif (s[j] == 'c'):
                cnt -= 1
 
            # If the frequency of 'a'
            # is greater than 'c'
            if (cnt > 0):
                ans+=1
 
    # Prthe answer
    print (ans)
 
# Drive Code
if __name__ == '__main__':
    S = "abccaab"
    countSubstrings(S)
 
# This code is contributed by mohit kumar 29.

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to find the number of
// substrings having the frequency of
// 'a' greater than frequency of 'c'
function countSubstrings(s)
{
    // Stores the size of the string
    var n = s.length;
 
    // Stores the resultant
    // count of substrings
    var ans = 0;
    var i,j;
    // Traverse the given string
    for (i = 0; i < n; i++) {
 
        // Store the difference between
        // frequency of 'a' and 'c'
        var cnt = 0;
 
        // Traverse all substrings
        // beginning at index i
        for (j = i; j < n; j++) {
            if (s[j] == 'a')
                cnt++;
            else if (s[j] == 'c')
                cnt--;
 
            // If the frequency of 'a'
            // is greater than 'c'
            if (cnt > 0) {
                ans++;
            }
        }
    }
 
    // Print the answer
    document.write(ans);
}
 
// Drive Code
    var S = "abccaab";
    countSubstrings(S);
 
</script>

C#




// C# program for the above approach
using System;
public class GFG {
 
    // Function to find the number of
    // substrings having the frequency of
    // 'a' greater than frequency of 'c'
    public static void countSubstrings(string s)
    {
 
        // Stores the size of the string
        int n = s.Length;
 
        // Stores the resultant
        // count of substrings
        int ans = 0;
 
        // Traverse the given string
        for (int i = 0; i < n; i++) {
 
            // Store the difference between
            // frequency of 'a' and 'c'
            int cnt = 0;
 
            // Traverse all substrings
            // beginning at index i
            for (int j = i; j < n; j++) {
                if (s[j] == 'a')
                    cnt++;
                else if (s[j] == 'c')
                    cnt--;
 
                // If the frequency of 'a'
                // is greater than 'c'
                if (cnt > 0) {
                    ans++;
                }
            }
        }
 
        // Print the answer
        Console.WriteLine(ans);
    }
 
    // Drive Code
    public static void Main(string[] args)
    {
        string S = "abccaab";
        countSubstrings(S);
    }
}
 
// This code is contributed by ukasp.
Output: 
11

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the Segment Tree. The idea is to store the difference of frequency of characters ‘a’ and ‘c’ for all prefixes of the string S in the segment tree node. Follow the steps below to solve the problem:

  • Initialize a variable, say count to 0, to store the difference between the frequency of characters ‘a’ and ‘c’.
  • Initialize a variable, say ans to 0, to store the count of substrings that have the frequency of character ‘a’ greater than ‘c’.
  • Initialize the segment tree with all 0s, which will be updated while traversing the string.
  • Since the difference between the frequency of characters ‘a’ and ‘c’ can be negative as well, all update operations on the segment tree will be done after adding N to the index that is to be updated to avoid negative indices.
  • Update the value of the index (0 + N) in the segment tree as the initial value of the count is 0.
  • Traverse the given string, S over the range [0, N – 1] and perform the following steps:
    • If the current character is ‘a’, then increment the count by 1. Otherwise, if the current character is ‘c’, then decrement the count by 1.
    • Perform the query on the segment tree to find the sum of all values less than count, as all these substrings will have the frequency of ‘a’ greater than ‘c’ and store the value returned in a variable say val.
    • Add the value of val to the variable ans.
    • Update the segment tree by incrementing the value at the index (count + N) by 1.
  • After completing the above steps, print the value of ans as the resultant count of substrings.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to update the segment Tree
void update(int ind, vector<int>& segTree,
            int n)
{
    // Update the value of ind
    ind += n;
 
    // Increment the leaf node
    segTree[ind]++;
 
    for (; ind > 1; ind >>= 1) {
 
        // Update the parent nodes
        segTree[ind >> 1] = segTree[ind]
                            + segTree[ind ^ 1];
    }
}
 
// Function to get the sum of all the
// elements between low to high - 1
int query(int low, int high,
          vector<int>& segTree, int n)
{
    // Initialize the leaf nodes of
    // the segment tree
    low += n;
    high += n;
 
    int ans = 0;
 
    while (low < high) {
 
        // Node lies completely in the
        // range of low to high
        if (low % 2) {
            ans += segTree[low];
            low++;
        }
 
        if (high % 2) {
            high--;
            ans += segTree[high];
        }
 
        // Update the value of nodes
        low >>= 1;
        high >>= 1;
    }
 
    return ans;
}
 
// Function to count the number of
// substrings which have frequency of
// 'a' greater than frequency of 'c'.
void countSubstrings(string& s)
{
    // Store the size of the string
    int n = s.length();
 
    // Initialize segment tree
    vector<int> segTree(4 * n);
 
    int count = 0;
 
    // Update the initial value of
    // the count
    update(n, segTree, 2 * n);
 
    // Stores the required result
    int ans = 0;
 
    // Traverse the given string
    for (int i = 0; i < n; i++) {
 
        // Increment count
        if (s[i] == 'a')
            count++;
 
        // Decrement count
        else if (s[i] == 'c')
            count--;
 
        // Query the segment tree to
        // find the sum of all values
        // less than count
        int val = query(0, n + count,
                        segTree, 2 * n);
        ans += val;
 
        // Update the current value of
        // count in the segment tree
        update(n + count, segTree, 2 * n);
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    string S = "abccaab";
    countSubstrings(S);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class Main
{
    static String s = "abccaab";
    static int n = s.length();
    static int[] segTree = new int[4*n];
      
    // Function to update the segment Tree
    static void update(int ind, int n)
    {
       
        // Update the value of ind
        ind += n;
   
        // Increment the leaf node
        segTree[ind]++;
   
        for (; ind > 1; ind >>= 1) {
   
            // Update the parent nodes
            segTree[ind >> 1] = segTree[ind]
                                + segTree[ind ^ 1];
        }
    }
   
    // Function to get the sum of all the
    // elements between low to high - 1
    static int query(int low, int high, int n)
    {
        
        // Initialize the leaf nodes of
        // the segment tree
        low += n;
        high += n;
   
        int ans = 0;
   
        while (low < high) {
   
            // Node lies completely in the
            // range of low to high
            if (low % 2 != 0) {
                ans += segTree[low];
                low++;
            }
   
            if (high % 2 != 0) {
                high--;
                ans += segTree[high];
            }
   
            // Update the value of nodes
            low >>= 1;
            high >>= 1;
        }
   
        return ans;
    }
      
    // Function to count the number of
    // substrings which have frequency of
    // 'a' greater than frequency of 'c'.
    static void countSubstrings()
    {
        int count = 0;
   
        // Update the initial value of
        // the count
        update(n, 2 * n);
   
        // Stores the required result
        int ans = 0;
   
        // Traverse the given string
        for (int i = 0; i < n; i++) {
   
            // Increment count
            if (s.charAt(i) == 'a')
                count++;
   
            // Decrement count
            else if (s.charAt(i) == 'c')
                count--;
   
            // Query the segment tree to
            // find the sum of all values
            // less than count
            int val = query(0, n + count, 2 * n);
            ans += val;
   
            // Update the current value of
            // count in the segment tree
            update(n + count, 2 * n);
        }
   
        // Print the answer
        System.out.print(ans);
    }
     
  // Driver code
    public static void main(String[] args) {
        Arrays.fill(segTree, 0);
        countSubstrings();
    }
}
 
// This code is contributed by mukesh07.

Python3




# Python3 program for the above approach
 
s = "abccaab"
n = len(s)
segTree = [0]*(4*n)
  
  
# Function to update the segment Tree
def update(ind, n):
 
    # Update the value of ind
    ind += n
 
    # Increment the leaf node
    segTree[ind]+=1
     
    while ind > 1:
        # Update the parent nodes
        segTree[ind >> 1] = segTree[ind] + segTree[ind ^ 1]
        ind >>= 1
 
# Function to get the sum of all the
# elements between low to high - 1
def query(low, high, n):
    # Initialize the leaf nodes of
    # the segment tree
    low += n
    high += n
 
    ans = 0
 
    while (low < high):
 
        # Node lies completely in the
        # range of low to high
        if (low % 2 != 0):
            ans += segTree[low]
            low+=1
 
        if (high % 2 != 0):
            high-=1
            ans += segTree[high]
 
        # Update the value of nodes
        low >>= 1
        high >>= 1
 
    return ans
 
# Function to count the number of
# substrings which have frequency of
# 'a' greater than frequency of 'c'.
def countSubstrings():
 
    count = 0
 
    # Update the initial value of
    # the count
    update(n, 2 * n)
 
    # Stores the required result
    ans = 0
 
    # Traverse the given string
    for i in range(n):
        # Increment count
        if (s[i] == 'a'):
            count+=1
 
        # Decrement count
        elif (s[i] == 'c'):
            count-=1
 
        # Query the segment tree to
        # find the sum of all values
        # less than count
        val = query(0, n + count, 2 * n)
        ans += val
 
        # Update the current value of
        # count in the segment tree
        update(n + count, 2 * n)
 
    # Print the answer
    print(ans)
 
countSubstrings()
 
# This code is contributed by suresh07.

C#




// C# program for the above approach
using System;
class GFG {
     
    static string s = "abccaab";
    static int n = s.Length;
    static int[] segTree = new int[4*n];
     
    // Function to update the segment Tree
    static void update(int ind, int n)
    {
        // Update the value of ind
        ind += n;
  
        // Increment the leaf node
        segTree[ind]++;
  
        for (; ind > 1; ind >>= 1) {
  
            // Update the parent nodes
            segTree[ind >> 1] = segTree[ind]
                                + segTree[ind ^ 1];
        }
    }
  
    // Function to get the sum of all the
    // elements between low to high - 1
    static int query(int low, int high, int n)
    {
       
        // Initialize the leaf nodes of
        // the segment tree
        low += n;
        high += n;
  
        int ans = 0;
  
        while (low < high) {
  
            // Node lies completely in the
            // range of low to high
            if (low % 2 != 0) {
                ans += segTree[low];
                low++;
            }
  
            if (high % 2 != 0) {
                high--;
                ans += segTree[high];
            }
  
            // Update the value of nodes
            low >>= 1;
            high >>= 1;
        }
  
        return ans;
    }
     
    // Function to count the number of
    // substrings which have frequency of
    // 'a' greater than frequency of 'c'.
    static void countSubstrings()
    {
        int count = 0;
  
        // Update the initial value of
        // the count
        update(n, 2 * n);
  
        // Stores the required result
        int ans = 0;
  
        // Traverse the given string
        for (int i = 0; i < n; i++) {
  
            // Increment count
            if (s[i] == 'a')
                count++;
  
            // Decrement count
            else if (s[i] == 'c')
                count--;
  
            // Query the segment tree to
            // find the sum of all values
            // less than count
            int val = query(0, n + count, 2 * n);
            ans += val;
  
            // Update the current value of
            // count in the segment tree
            update(n + count, 2 * n);
        }
  
        // Print the answer
        Console.Write(ans);
    }
     
  // Driver code
  static void Main() {
    Array.Fill(segTree, 0);
    countSubstrings();
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
    // Javascript program for the above approach
     
    s = "abccaab";
    n = s.length
    segTree = new Array(4*n);
    segTree.fill(0);
     
     
    // Function to update the segment Tree
    function update(ind, n)
    {
        // Update the value of ind
        ind += n;
 
        // Increment the leaf node
        segTree[ind]++;
 
        for (; ind > 1; ind >>= 1) {
 
            // Update the parent nodes
            segTree[ind >> 1] = segTree[ind]
                                + segTree[ind ^ 1];
        }
    }
 
    // Function to get the sum of all the
    // elements between low to high - 1
    function query(low, high, n)
    {
        // Initialize the leaf nodes of
        // the segment tree
        low += n;
        high += n;
 
        let ans = 0;
 
        while (low < high) {
 
            // Node lies completely in the
            // range of low to high
            if (low % 2) {
                ans += segTree[low];
                low++;
            }
 
            if (high % 2) {
                high--;
                ans += segTree[high];
            }
 
            // Update the value of nodes
            low >>= 1;
            high >>= 1;
        }
 
        return ans;
    }
 
    // Function to count the number of
    // substrings which have frequency of
    // 'a' greater than frequency of 'c'.
    function countSubstrings()
    {
        let count = 0;
 
        // Update the initial value of
        // the count
        update(n, 2 * n);
 
        // Stores the required result
        let ans = 0;
 
        // Traverse the given string
        for (let i = 0; i < n; i++) {
 
            // Increment count
            if (s[i] == 'a')
                count++;
 
            // Decrement count
            else if (s[i] == 'c')
                count--;
 
            // Query the segment tree to
            // find the sum of all values
            // less than count
            let val = query(0, n + count, 2 * n);
            ans += val;
 
            // Update the current value of
            // count in the segment tree
            update(n + count, 2 * n);
        }
 
        // Print the answer
        document.write(ans);
    }
     
    countSubstrings();
    
   // This code is contributed by decode2207.
</script>
Output: 
11

 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :