Count substrings having frequency of a character exceeding that of another character in a string
Last Updated :
15 Jul, 2025
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:
- "a": The frequency of a and c is 1 and 0 respectively.
- "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 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
# 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
# Print the answer
print (ans)
# Drive Code
if __name__ == '__main__':
S = "abccaab"
countSubstrings(S)
# This code is contributed by mohit kumar 29.
<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# 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.
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 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 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# 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.
<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>
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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