# Index of character depending on frequency count in string

• Last Updated : 08 Jun, 2021

Given a string str containing only lowercase characters, the task is to answer Q queries of the following type:

1. 1 C X: Find the largest i such that str[0…i] has exactly X occurrence of the character C.
2. 2 C X: Find the smallest i such that str[0…i] has exactly X occurrence of the character C.

Example:

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.

Input: str = “geeksforgeeks”, query[] = {{1, ‘e’, 2}, {2, ‘k’, 2}}
Output:

11
Query 1: “geeksforg” is the largest substring starting at str with ‘e’ appearing exactly twice and the index of the last character is 8.
Query 2: “geeksforgeek” is the smallest substring starting at str with ‘k’ appearing exactly twice and the index of the last character is 11.
Input: str = “abcdabcd”, query[] = {{1, ‘a’, 1}, {2, ‘a’, 2}}
Output:

Approach: Create two 2-dimensional arrays L[][] and F[][] such that L[i][j] stores the largest i such that the ith character appears exactly jth times in str[0…i] and F[i][j] stores the smallest i such that the ith character appears exactly jth times in str[0…i]. In order to do so, traverse the whole string and maintain a frequency array so that for each iteration/character, its count is updated and then start another loop from 0 to 26 (each letter a-z). In the inner loop, if the iterator is equal to character value then update L[][] and F[][] array with the current index position using outer loop iterator otherwise just increment the L[][] array value for other characters by 1 as only index has been incremented and the character has not occurred. Now, type 1 query can be answered as L[given character][Frequency count] and type 2 query as F[given character][Frequency count].
Below is the implementation of the above approach.

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `const` `int` `MAX = 26;` `// Function to perform the queries``void` `performQueries(string str, ``int` `q, ``int` `type[],``                    ``char` `ch[], ``int` `freq[])``{` `    ``int` `n = str.length();` `    ``// L[i][j] stores the largest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `L[MAX][n];` `    ``// F[i][j] stores the smallest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `F[MAX][n];` `    ``// To store the frequency of each``    ``// of the character of str``    ``int` `cnt[MAX] = { 0 };``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Current character of str``        ``int` `k = str[i] - ``'a'``;` `        ``// Update its frequency``        ``cnt[k]++;` `        ``// For every lowercase character``        ``// of the English alphabet``        ``for` `(``int` `j = 0; j < MAX; j++) {` `            ``// If it is equal to the character``            ``// under consideration then update``            ``// L[][] and R[][] as it is cnt[j]th``            ``// occurrence of character k``            ``if` `(k == j) {``                ``L[j][cnt[j]] = i;``                ``F[j][cnt[j]] = i;``            ``}` `            ``// Only update L[][] as k has not``            ``// been occurred so only index``            ``// has to be incremented``            ``else``                ``L[j][cnt[j]] = L[j][cnt[j]] + 1;``        ``}``    ``}` `    ``// Perform the queries``    ``for` `(``int` `i = 0; i < q; i++) {` `        ``// Type 1 query``        ``if` `(type[i] == 1) {``            ``cout << L[ch[i] - ``'a'``][freq[i]];``        ``}` `        ``// Type 2 query``        ``else` `{``            ``cout << F[ch[i] - ``'a'``][freq[i]];``        ``}` `        ``cout << ``"\n"``;``    ``}``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;` `    ``// Queries``    ``int` `type[] = { 1, 2 };``    ``char` `ch[] = { ``'e'``, ``'k'` `};``    ``int` `freq[] = { 2, 2 };``    ``int` `q = ``sizeof``(type) / ``sizeof``(``int``);` `    ``// Perform the queries``    ``performQueries(str, q, type, ch, freq);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``static` `int` `MAX = ``26``;` `// Function to perform the queries``static` `void` `performQueries(String str, ``int` `q, ``int` `type[],``                                   ``char` `ch[], ``int` `freq[])``{``    ``int` `n = str.length();` `    ``// L[i][j] stores the largest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `[][]L = ``new` `int``[MAX][n];` `    ``// F[i][j] stores the smallest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `[][]F = ``new` `int``[MAX][n];` `    ``// To store the frequency of each``    ``// of the character of str``    ``int` `[]cnt = ``new` `int``[MAX];``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Current character of str``        ``int` `k = str.charAt(i) - ``'a'``;` `        ``// Update its frequency``        ``cnt[k]++;` `        ``// For every lowercase character``        ``// of the English alphabet``        ``for` `(``int` `j = ``0``; j < MAX; j++)``        ``{` `            ``// If it is equal to the character``            ``// under consideration then update``            ``// L[][] and R[][] as it is cnt[j]th``            ``// occurrence of character k``            ``if` `(k == j)``            ``{``                ``L[j][cnt[j]] = i;``                ``F[j][cnt[j]] = i;``            ``}` `            ``// Only update L[][] as k has not``            ``// been occurred so only index``            ``// has to be incremented``            ``else``                ``L[j][cnt[j]] = L[j][cnt[j]] + ``1``;``        ``}``    ``}` `    ``// Perform the queries``    ``for` `(``int` `i = ``0``; i < q; i++)``    ``{` `        ``// Type 1 query``        ``if` `(type[i] == ``1``)``        ``{``            ``System.out.print(L[ch[i] - ``'a'``][freq[i]]);``        ``}` `        ``// Type 2 query``        ``else``        ``{``            ``System.out.print(F[ch[i] - ``'a'``][freq[i]]);``        ``}``        ``System.out.print(``"\n"``);``    ``}``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``String str = ``"geeksforgeeks"``;` `    ``// Queries``    ``int` `type[] = { ``1``, ``2` `};``    ``char` `ch[] = { ``'e'``, ``'k'` `};``    ``int` `freq[] = { ``2``, ``2` `};``    ``int` `q = type.length;` `    ``// Perform the queries``    ``performQueries(str, q, type, ch, freq);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach``import` `numpy as np` `MAX` `=` `26``;` `# Function to perform the queries``def` `performQueries(string , q, type_arr, ch, freq) :` `    ``n ``=` `len``(string);` `    ``# L[i][j] stores the largest i``    ``# such that ith character appears``    ``# exactly jth times in str[0...i]``    ``L ``=` `np.zeros((``MAX``, n));` `    ``# F[i][j] stores the smallest i``    ``# such that ith character appears``    ``# exactly jth times in str[0...i]``    ``F ``=` `np.zeros((``MAX``, n));` `    ``# To store the frequency of each``    ``# of the character of str``    ``cnt ``=` `[ ``0` `] ``*` `MAX``;``    ``for` `i ``in` `range``(n) :` `        ``# Current character of str``        ``k ``=` `ord``(string[i]) ``-` `ord``(``'a'``);` `        ``# Update its frequency``        ``cnt[k] ``+``=` `1``;` `        ``# For every lowercase character``        ``# of the English alphabet``        ``for` `j ``in` `range``(``MAX``) :` `            ``# If it is equal to the character``            ``# under consideration then update``            ``# L[][] and R[][] as it is cnt[j]th``            ``# occurrence of character k``            ``if` `(k ``=``=` `j) :``                ``L[j][cnt[j]] ``=` `i;``                ``F[j][cnt[j]] ``=` `i;` `            ``# Only update L[][] as k has not``            ``# been occurred so only index``            ``# has to be incremented``            ``else` `:``                ``L[j][cnt[j]] ``=` `L[j][cnt[j]] ``+` `1``;` `    ``# Perform the queries``    ``for` `i ``in` `range``(q) :` `        ``# Type 1 query``        ``if` `(type_arr[i] ``=``=` `1``) :``            ``print``(L[``ord``(ch[i]) ``-``                    ``ord``(``'a'``)][freq[i]], end ``=` `"");` `        ``# Type 2 query``        ``else` `:``            ``print``(F[``ord``(ch[i]) ``-``                    ``ord``(``'a'``)][freq[i]], end ``=` `"");``            ` `        ``print``()``        ` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``string ``=` `"geeksforgeeks"``;` `    ``# Queries``    ``type_arr ``=` `[ ``1``, ``2` `];``    ``ch ``=` `[ ``'e'``, ``'k'` `];``    ``freq ``=` `[ ``2``, ``2` `];``    ``q ``=` `len``(type_arr);` `    ``# Perform the queries``    ``performQueries(string, q, type_arr, ch, freq);` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``static` `int` `MAX = 26;` `// Function to perform the queries``static` `void` `performQueries(String str, ``int` `q, ``int` `[]type,``                                     ``char` `[]ch, ``int` `[]freq)``{``    ``int` `n = str.Length;` `    ``// L[i,j] stores the largest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `[,]L = ``new` `int``[MAX, n];` `    ``// F[i,j] stores the smallest i``    ``// such that ith character appears``    ``// exactly jth times in str[0...i]``    ``int` `[,]F = ``new` `int``[MAX, n];` `    ``// To store the frequency of each``    ``// of the character of str``    ``int` `[]cnt = ``new` `int``[MAX];``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Current character of str``        ``int` `k = str[i] - ``'a'``;` `        ``// Update its frequency``        ``cnt[k]++;` `        ``// For every lowercase character``        ``// of the English alphabet``        ``for` `(``int` `j = 0; j < MAX; j++)``        ``{` `            ``// If it is equal to the character``            ``// under consideration then update``            ``// L[,] and R[,] as it is cnt[j]th``            ``// occurrence of character k``            ``if` `(k == j)``            ``{``                ``L[j, cnt[j]] = i;``                ``F[j, cnt[j]] = i;``            ``}` `            ``// Only update L[,] as k has not``            ``// been occurred so only index``            ``// has to be incremented``            ``else``                ``L[j, cnt[j]] = L[j, cnt[j]] + 1;``        ``}``    ``}` `    ``// Perform the queries``    ``for` `(``int` `i = 0; i < q; i++)``    ``{` `        ``// Type 1 query``        ``if` `(type[i] == 1)``        ``{``            ``Console.Write(L[ch[i] - ``'a'``, freq[i]]);``        ``}` `        ``// Type 2 query``        ``else``        ``{``            ``Console.Write(F[ch[i] - ``'a'``, freq[i]]);``        ``}``        ``Console.Write(``"\n"``);``    ``}``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``String str = ``"geeksforgeeks"``;` `    ``// Queries``    ``int` `[]type = { 1, 2 };``    ``char` `[]ch = { ``'e'``, ``'k'` `};``    ``int` `[]freq = { 2, 2 };``    ``int` `q = type.Length;` `    ``// Perform the queries``    ``performQueries(str, q, type, ch, freq);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
```8
11```

Time Complexity: O(n) where n is the length of the string.

My Personal Notes arrow_drop_up