# Count of lexicographically smaller characters on right

Given a string consisting of only lowercase English alphabets. The task is to count the total number of alphabetically smaller characters on the right side of characters at each index.

Examples:

Input: str = “edcba”
Output: 4 3 2 1 0
Explanation:
Number of characters on the right side of index 0 which is smaller than
e are dcba = 4
Number of characters on the right side of index 1 which is smaller than
d are cba = 3
Number of characters on the right side of index 2 which is smaller than
c are ba = 2
Number of characters on the right side of index 3 which is smaller than
b are a = 1
Number of characters on the right side of index 4 which is smaller than
a are ‘\0’ = 0
Input: str = “eaaa”
Output: 3 0 0 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:
The idea is to traverse all the characters on the right side of each index of string one by one and print the count of characters which are alphabetically smaller.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// function to count the smaller ` `// characters at the right of index i ` `void` `countSmaller(string str) ` `{ ` ` `  `    ``// store the length of string ` `    ``int` `n = str.length(); ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// for each index initialize ` `        ``// count as zero ` `        ``int` `cnt = 0; ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` ` `  `            ``// increment the count if ` `            ``// characters are smaller ` `            ``// than at ith index ` `            ``if` `(str[j] < str[i]) { ` `                ``cnt += 1; ` `            ``} ` `        ``} ` ` `  `        ``// print the count of characters ` `        ``// smaller than the index i ` `        ``cout << cnt << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// input string ` `    ``string str = ``"edcba"``; ` `    ``countSmaller(str); ` `} `

## Java

 `class` `GFG ` `{ ` ` `  `// function to count the smaller ` `// characters at the right of index i ` `static` `void` `countSmaller(String str) ` `{ ` ` `  `    ``// store the length of String ` `    ``int` `n = str.length(); ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// for each index initialize ` `        ``// count as zero ` `        ``int` `cnt = ``0``; ` `        ``for` `(``int` `j = i + ``1``; j < n; j++) ` `        ``{ ` ` `  `            ``// increment the count if ` `            ``// characters are smaller ` `            ``// than at ith index ` `            ``if` `(str.charAt(j) < str.charAt(i)) ` `            ``{ ` `                ``cnt += ``1``; ` `            ``} ` `        ``} ` ` `  `        ``// print the count of characters ` `        ``// smaller than the index i ` `        ``System.out.print(cnt+ ``" "``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// input String ` `    ``String str = ``"edcba"``; ` `    ``countSmaller(str); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# function to count the smaller ` `# characters at the right of index i ` `def` `countSmaller(``str``): ` ` `  `    ``# store the length of String ` `    ``n ``=` `len``(``str``); ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# for each index initialize ` `        ``# count as zero ` `        ``cnt ``=` `0``; ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` ` `  `            ``# increment the count if ` `            ``# characters are smaller ` `            ``# than at ith index ` `            ``if` `(``str``[j] < ``str``[i]): ` `                ``cnt ``+``=` `1``; ` `             `  `        ``# print the count of characters ` `        ``# smaller than the index i ` `        ``print``(cnt, end ``=``" "``); ` `     `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``# input String ` `    ``str` `=` `"edcba"``; ` `    ``countSmaller(``str``); ` ` `  `# This code is contributed by PrinciRaj1992 `

## C#

 `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// function to count the smaller ` `// characters at the right of index i ` `static` `void` `countSmaller(String str) ` `{ ` ` `  `    ``// store the length of String ` `    ``int` `n = str.Length; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// for each index initialize ` `        ``// count as zero ` `        ``int` `cnt = 0; ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `        ``{ ` ` `  `            ``// increment the count if ` `            ``// characters are smaller ` `            ``// than at ith index ` `            ``if` `(str[j] < str[i]) ` `            ``{ ` `                ``cnt += 1; ` `            ``} ` `        ``} ` ` `  `        ``// print the count of characters ` `        ``// smaller than the index i ` `        ``Console.Write(cnt+ ``" "``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// input String ` `    ``String str = ``"edcba"``; ` `    ``countSmaller(str); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```4 3 2 1 0
```

Time Complexity: O(N2), where N = length of string
Auxiliary Space: O(1)

Better Approach: The idea is to Use Self Balancing BST.
A Self Balancing Binary Search Tree (AVL, Red Black, .. etc) can be used to get the solution in O(N log N) time complexity. We can augment these trees so that every node N contains size the subtree rooted with N. We have used AVL tree in the following implementation.
We traverse the string from right to left and insert all elements one by one in an AVL tree. While inserting a new key in an AVL tree, we first compare the key with root. If the key is greater than root, then it is greater than all the nodes in the left subtree of root. So we add the size of the left subtree to the count of smaller elements for the key being inserted. We recursively follow the same approach for all nodes down the root.

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

Efficient Approach:
The idea is to use hashing technique because string consists of only lowercase alphabets. So here we can take an array of size 26 that is used to store the count of smaller characters on the right side of that index.

Traverse the array from right and keep updating the count of characters in the hash array. Now, to find the count of characters smaller on right, we can simply traverse the hash array of size 26 every time to count smaller characters encountered so far.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Function to count the smaller ` `// characters on the right of index i ` `void` `countSmaller(string str) ` `{ ` `    ``// store the length of string ` `    ``int` `n = str.length(); ` ` `  `    ``// initialize each elements of ` `    ``// arr to zero ` `    ``int` `arr[26] = { 0 }; ` ` `  `    ``// array to store count of smaller characters on ` `    ``// the right side of that index ` `    ``int` `ans[n]; ` ` `  `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` ` `  `        ``arr[str[i] - ``'a'``]++; ` ` `  `        ``// initialize the variable to store ` `        ``// the count of characters smaller ` `        ``// than that at index i ` `        ``int` `ct = 0; ` ` `  `        ``// adding the count of characters ` `        ``// smaller than index i ` `        ``for` `(``int` `j = 0; j < str[i] - ``'a'``; j++) { ` `            ``ct += arr[j]; ` `        ``} ` `        ``ans[i] = ct; ` `    ``} ` ` `  `    ``// print the count of characters smaller ` `    ``// than index i stored in ans array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << ans[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string str = ``"edcbaa"``; ` `    ``countSmaller(str); ` ` `  `    ``return` `0; ` `} `

## Java

 `class` `GFG ` `{ ` ` `  `// Function to count the smaller ` `// characters on the right of index i ` `static` `void` `countSmaller(String str) ` `{ ` `    ``// store the length of String ` `    ``int` `n = str.length(); ` ` `  `    ``// initialize each elements of ` `    ``// arr to zero ` `    ``int` `arr[] = ``new` `int``[``26``]; ` ` `  `    ``// array to store count of smaller characters on ` `    ``// the right side of that index ` `    ``int` `ans[] = ``new` `int``[n]; ` ` `  `    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) ` `    ``{ ` ` `  `        ``arr[str.charAt(i) - ``'a'``]++; ` ` `  `        ``// initialize the variable to store ` `        ``// the count of characters smaller ` `        ``// than that at index i ` `        ``int` `ct = ``0``; ` ` `  `        ``// adding the count of characters ` `        ``// smaller than index i ` `        ``for` `(``int` `j = ``0``; j < str.charAt(i) - ``'a'``; j++)  ` `        ``{ ` `            ``ct += arr[j]; ` `        ``} ` `        ``ans[i] = ct; ` `    ``} ` ` `  `    ``// print the count of characters smaller ` `    ``// than index i stored in ans array ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``System.out.print(ans[i] + ``" "``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"edcbaa"``; ` `    ``countSmaller(str); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Function to count the smaller ` `# characters on the right of index i ` `def` `countSmaller(``str``): ` ` `  `    ``# store the length of String ` `    ``n ``=` `len``(``str``); ` ` `  `    ``# initialize each elements of ` `    ``# arr to zero ` `    ``arr ``=` `[``0``]``*``26``; ` ` `  `    ``# array to store count of smaller characters on ` `    ``# the right side of that index ` `    ``ans ``=` `[``0``]``*``n; ` ` `  `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``): ` ` `  `        ``arr[``ord``(``str``[i] ) ``-` `ord``(``'a'``)] ``+``=` `1``; ` ` `  `        ``# initialize the variable to store ` `        ``# the count of characters smaller ` `        ``# than that at index i ` `        ``ct ``=` `0``; ` ` `  `        ``# adding the count of characters ` `        ``# smaller than index i ` `        ``for` `j ``in` `range``(``ord``(``str``[i] ) ``-` `ord``(``'a'``)): ` `            ``ct ``+``=` `arr[j]; ` `         `  `        ``ans[i] ``=` `ct; ` `     `  `    ``# print the count of characters smaller ` `    ``# than index i stored in ans array ` `    ``for` `i ``in` `range``(n): ` `        ``print``(ans[i], end ``=` `" "``); ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``str` `=` `"edcbaa"``; ` `    ``countSmaller(``str``); ` `     `  `# This code is contributed by Rajput-Ji `

## C#

 `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to count the smaller ` `// characters on the right of index i ` `static` `void` `countSmaller(String str) ` `{ ` `    ``// store the length of String ` `    ``int` `n = str.Length; ` ` `  `    ``// initialize each elements of ` `    ``// arr to zero ` `    ``int` `[]arr = ``new` `int``[26]; ` ` `  `    ``// array to store count of smaller characters on ` `    ``// the right side of that index ` `    ``int` `[]ans = ``new` `int``[n]; ` ` `  `    ``for` `(``int` `i = n - 1; i >= 0; i--) ` `    ``{ ` ` `  `        ``arr[str[i] - ``'a'``]++; ` ` `  `        ``// initialize the variable to store ` `        ``// the count of characters smaller ` `        ``// than that at index i ` `        ``int` `ct = 0; ` ` `  `        ``// adding the count of characters ` `        ``// smaller than index i ` `        ``for` `(``int` `j = 0; j < str[i] - ``'a'``; j++)  ` `        ``{ ` `            ``ct += arr[j]; ` `        ``} ` `        ``ans[i] = ct; ` `    ``} ` ` `  `    ``// print the count of characters smaller ` `    ``// than index i stored in ans array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``Console.Write(ans[i] + ``" "``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String str = ``"edcbaa"``; ` `    ``countSmaller(str); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```5 4 3 2 0 0
```

Time Complexity: O(N*26)
Auxiliary Space: O(N+26)

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