# Minimum letters to be removed to make all occurrences of a given letter continuous

• Last Updated : 29 Aug, 2022

Given a string str and a character K, the task is to find the minimum number of elements that should be removed from this sequence so that all occurrences of the given character K become continuous.

Examples:

Input: str = “ababababa”, K = ‘a’
Output:
Explanation:
All the occurrences of the character ‘b’ should be removed in order to make the occurrences of ‘a’ continuous.

Input: str = “kprkkoinkopt”, K = ‘k’
Output:

Approach: The idea is to find the number of characters other than K in between the first and last occurrence of the given character K. In order to do this, the following steps are followed:

1. Find the first occurrence of the character K.
2. Find the last occurrence of the character K.
3. Iterate between the indices and find the number of characters present in between those indices other than K. This is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum number of``// deletions required to make the occurrences``// of the given character K continuous``int` `noOfDeletions(string str, ``char` `k)``{``    ``int` `ans = 0, cnt = 0, pos = 0;` `    ``// Find the first occurrence of the given letter``    ``while` `(pos < str.length() && str[pos] != k) {``        ``pos++;``    ``}` `    ``int` `i = pos;` `    ``// Iterate from the first occurrence``    ``// till the end of the sequence``    ``while` `(i < str.length()) {` `        ``// Find the index from where the occurrence``        ``// of the character is not continuous``        ``while` `(i < str.length() && str[i] == k) {``            ``i = i + 1;``        ``}` `        ``// Update the answer with the number of``        ``// elements between non-consecutive occurrences``        ``// of the given letter``        ``ans = ans + cnt;``        ``cnt = 0;``        ``while` `(i < str.length() && str[i] != k) {``            ``i = i + 1;` `            ``// Update the count for all letters``            ``// which are not equal to the given letter``            ``cnt = cnt + 1;``        ``}``    ``}` `    ``// Return the count``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``string str1 = ``"ababababa"``;``    ``char` `k1 = ``'a'``;``    ``// Calling the function``    ``cout << noOfDeletions(str1, k1) << endl;` `    ``string str2 = ``"kprkkoinkopt"``;``    ``char` `k2 = ``'k'``;``    ``// Calling the function``    ``cout << noOfDeletions(str2, k2) << endl;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;` `class` `GFG{`` ` `// Function to find the minimum number of``// deletions required to make the occurrences``// of the given character K continuous``static` `int` `noOfDeletions(String str, ``char` `k)``{``    ``int` `ans = ``0``, cnt = ``0``, pos = ``0``;`` ` `    ``// Find the first occurrence of the given letter``    ``while` `(pos < str.length() && str.charAt(pos) != k) {``        ``pos++;``    ``}`` ` `    ``int` `i = pos;`` ` `    ``// Iterate from the first occurrence``    ``// till the end of the sequence``    ``while` `(i < str.length()) {`` ` `        ``// Find the index from where the occurrence``        ``// of the character is not continuous``        ``while` `(i < str.length() && str.charAt(i) == k) {``            ``i = i + ``1``;``        ``}`` ` `        ``// Update the answer with the number of``        ``// elements between non-consecutive occurrences``        ``// of the given letter``        ``ans = ans + cnt;``        ``cnt = ``0``;``        ``while` `(i < str.length() && str.charAt(i) != k) {``            ``i = i + ``1``;`` ` `            ``// Update the count for all letters``            ``// which are not equal to the given letter``            ``cnt = cnt + ``1``;``        ``}``    ``}`` ` `    ``// Return the count``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String str1 = ``"ababababa"``;``    ``char` `k1 = ``'a'``;` `    ``// Calling the function``    ``System.out.print(noOfDeletions(str1, k1) +``"\n"``);`` ` `    ``String str2 = ``"kprkkoinkopt"``;``    ``char` `k2 = ``'k'``;` `    ``// Calling the function``    ``System.out.print(noOfDeletions(str2, k2) +``"\n"``);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of the above approach` `# Function to find the minimum number of``# deletions required to make the occurrences``# of the given character K continuous``def` `noOfDeletions(string, k) :` `    ``ans ``=` `0``; cnt ``=` `0``; pos ``=` `0``;` `    ``# Find the first occurrence of the given letter``    ``while` `(pos < ``len``(string) ``and` `string[pos] !``=` `k) :``        ``pos ``+``=` `1``;` `    ``i ``=` `pos;` `    ``# Iterate from the first occurrence``    ``# till the end of the sequence``    ``while` `(i < ``len``(string)) :` `        ``# Find the index from where the occurrence``        ``# of the character is not continuous``        ``while` `(i < ``len``(string) ``and` `string[i] ``=``=` `k) :``            ``i ``=` `i ``+` `1``;``        ` `        ``# Update the answer with the number of``        ``# elements between non-consecutive occurrences``        ``# of the given letter``        ``ans ``=` `ans ``+` `cnt;``        ``cnt ``=` `0``;``        ``while` `(i < ``len``(string) ``and` `string[i] !``=` `k) :``            ``i ``=` `i ``+` `1``;` `            ``# Update the count for all letters``            ``# which are not equal to the given letter``            ``cnt ``=` `cnt ``+` `1``;``            ` `    ``# Return the count``    ``return` `ans;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``str1 ``=` `"ababababa"``;``    ``k1 ``=` `'a'``;``    ` `    ``# Calling the function``    ``print``(noOfDeletions(str1, k1));``    ` `    ``str2 ``=` `"kprkkoinkopt"``;``    ``k2 ``=` `'k'``;``    ` `    ``# Calling the function``    ``print``(noOfDeletions(str2, k2));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``  ` `// Function to find the minimum number of``// deletions required to make the occurrences``// of the given character K continuous``static` `int` `noOfDeletions(String str, ``char` `k)``{``    ``int` `ans = 0, cnt = 0, pos = 0;``  ` `    ``// Find the first occurrence of the given letter``    ``while` `(pos < str.Length && str[pos] != k) {``        ``pos++;``    ``}``  ` `    ``int` `i = pos;``  ` `    ``// Iterate from the first occurrence``    ``// till the end of the sequence``    ``while` `(i < str.Length) {``  ` `        ``// Find the index from where the occurrence``        ``// of the character is not continuous``        ``while` `(i < str.Length && str[i] == k) {``            ``i = i + 1;``        ``}``  ` `        ``// Update the answer with the number of``        ``// elements between non-consecutive occurrences``        ``// of the given letter``        ``ans = ans + cnt;``        ``cnt = 0;``        ``while` `(i < str.Length && str[i] != k) {``            ``i = i + 1;``  ` `            ``// Update the count for all letters``            ``// which are not equal to the given letter``            ``cnt = cnt + 1;``        ``}``    ``}``  ` `    ``// Return the count``    ``return` `ans;``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String str1 = ``"ababababa"``;``    ``char` `k1 = ``'a'``;`` ` `    ``// Calling the function``    ``Console.Write(noOfDeletions(str1, k1) +``"\n"``);``  ` `    ``String str2 = ``"kprkkoinkopt"``;``    ``char` `k2 = ``'k'``;`` ` `    ``// Calling the function``    ``Console.Write(noOfDeletions(str2, k2) +``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

```4
5```

Time Complexity:  O(N) since one traversal of the string is required to complete all operations hence the overall time required by the algorithm is linear

Auxiliary Space: O(1) since no extra array is used the space taken by the algorithm is linear.

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