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:4Explanation:

All the occurances of the character ‘b’ should be removed in order to make the occurances of ‘a’ continuous.

Input:str = “kprkkoinkopt”, K = ‘k’Output:5

**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:

- Find the first occurrence of the character
**K**. - Find the last occurrence of the character
**K**. - 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 <bits/stdc++.h>` `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` |

**Output:**

4 5

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