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

• Difficulty Level : Hard
• Last Updated : 29 Jan, 2022

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 ` `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`   `    ``# Print the answer` `    ``print` `(ans)`   `# Drive Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``S ``=` `"abccaab"` `    ``countSubstrings(S)`   `# This code is contributed by mohit kumar 29.`

## Javascript

 ``

## 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 ` `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

 ``

Output:

`11`

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

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