# Longest suffix such that occurrence of each character is less than N after deleting atmost K characters

Given a string S and two integer N and K, the task is to find the maximum length suffix such that occurrence of each character in the suffix string is less than N and atmost K elements can be deleted from the input string to get maximum length suffix.

Examples:

Input: S = “iahagafedcba”, N = 1, K = 0
Output: fedcba
Explanation:
Maximum length suffix in the given string
Such that occurence of each character is 1 is “fedcba”,
Because if we take the string “afedcba”,
then the occurence of character “a” will be 2.

Input: S = “iahagafedcba”, N = 1, K = 2
Output: hgfedcba
Explanation:
Maximum length suffix in the given string
Such that occurrence of each character is 1 is “hgfedcba”,
After deleting character “a” two times.

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

Approach: The idea is to use hash-map to store the frequency of the characters of the string.

• Intialize an empty string to store the longest suffix of the string.
• Iterate the string from last using a loop –
• Increment the occurrence of the character by 1 in the hash-map
• If the occurrence of current character is less than N, then add the character into the suffix string
• Otherwise, decrement the value of K by 1 if the value of K is greater than 0
• Print the suffix string.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find ` `// longest suffix of the string ` `// such that occurrence of each  ` `// character is less than K ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum ` `// length suffix in the string ` `void` `maximumSuffix(string s,  ` `               ``int` `n, ``int` `k){ ` `     `  `    ``// Length of the string ` `    ``int` `i = s.length() - 1; ` ` `  `    ``// Map to store the number ` `    ``// of occurrence of character ` `    ``int` `arr = { 0 }; ` `    ``string suffix = ``""``; ` ` `  `    ``// Loop to iterate string  ` `    ``// from the last character ` `    ``while` `(i > -1) { ` ` `  `        ``int` `index = s[i] - ``'a'``; ` `         `  `        ``// Condition to check if the  ` `        ``// occurrence of each character ` `        ``// is less than given number ` `        ``if` `(arr[index] < n) { ` `            ``arr[index]++; ` `            ``suffix += s[i]; ` `            ``i--; ` `            ``continue``; ` `        ``} ` `         `  `        ``// Condition when character ` `        ``// cannot be deleted ` `        ``if` `(k == 0) ` `            ``break``; ` `        ``k--; ` `        ``i--; ` `    ``} ` `    ``reverse(suffix.begin(), suffix.end()); ` ` `  `    ``// Longest suffix ` `    ``cout << suffix; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string str = ``"iahagafedcba"``; ` ` `  `    ``int` `n = 1, k = 2; ` `     `  `    ``// Function call ` `    ``maximumSuffix(str, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find ` `// longest suffix of the String ` `// such that occurrence of each  ` `// character is less than K ` `class` `GFG{ ` ` `  `// Function to find the maximum ` `// length suffix in the String ` `static` `void` `maximumSuffix(String s,  ` `            ``int` `n, ``int` `k){ ` `     `  `    ``// Length of the String ` `    ``int` `i = s.length() - ``1``; ` ` `  `    ``// Map to store the number ` `    ``// of occurrence of character ` `    ``int` `arr[] = ``new` `int``[``26``]; ` `    ``String suffix = ``""``; ` `     `  `    ``// Loop to iterate String  ` `    ``// from the last character ` `    ``while` `(i > -``1``) { ` ` `  `        ``int` `index = s.charAt(i) - ``'a'``; ` `         `  `        ``// Condition to check if the  ` `        ``// occurrence of each character ` `        ``// is less than given number ` `        ``if` `(arr[index] < n) { ` `            ``arr[index]++; ` `            ``suffix += s.charAt(i); ` `            ``i--; ` `            ``continue``; ` `        ``} ` `         `  `        ``// Condition when character ` `        ``// cannot be deleted ` `        ``if` `(k == ``0``) ` `            ``break``; ` `        ``k--; ` `        ``i--; ` `    ``} ` `    ``suffix = reverse(suffix); ` `     `  `    ``// Longest suffix ` `    ``System.out.print(suffix); ` `} ` `static` `String reverse(String input) { ` `    ``char``[] a = input.toCharArray(); ` `    ``int` `l, r = a.length - ``1``; ` `    ``for` `(l = ``0``; l < r; l++, r--) { ` `        ``char` `temp = a[l]; ` `        ``a[l] = a[r]; ` `        ``a[r] = temp; ` `    ``} ` `    ``return` `String.valueOf(a); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"iahagafedcba"``; ` ` `  `    ``int` `n = ``1``, k = ``2``; ` `     `  `    ``// Function call ` `    ``maximumSuffix(str, n, k); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python 3

 `# Python 3 implementation to find ` `# longest suffix of the string ` `# such that occurrence of each  ` `# character is less than K ` ` `  `# Function to find the maximum ` `# length suffix in the string ` `def` `maximumSuffix(s, n, k): ` `     `  `    ``# Length of the string ` `    ``i ``=` `len``(s)``-` `1` ` `  `    ``# Map to store the number ` `    ``# of occurrence of character ` `    ``arr ``=` `[``0` `for` `i ``in` `range``(``26``)] ` `    ``suffix ``=` `"" ` `     `  `    ``# Loop to iterate string  ` `    ``# from the last character ` `    ``while` `(i > ``-``1``): ` `        ``index ``=` `ord``(s[i]) ``-` `ord``(``'a'``); ` `         `  `        ``# Condition to check if the  ` `        ``# occurrence of each character ` `        ``# is less than given number ` `        ``if` `(arr[index] < n): ` `            ``arr[index] ``+``=` `1` `            ``suffix ``+``=` `s[i] ` `            ``i ``-``=` `1` `            ``continue` `         `  `        ``# Condition when character ` `        ``# cannot be deleted ` `        ``if` `(k ``=``=` `0``): ` `            ``break` `        ``k ``-``=` `1` `        ``i ``-``=` `1` `     `  `    ``suffix ``=` `suffix[::``-``1``] ` `     `  `    ``# Longest suffix ` `    ``print``(suffix) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``str` `=` `"iahagafedcba"` ` `  `    ``n ``=` `1` `    ``k ``=` `2` `     `  `    ``# Function call ` `    ``maximumSuffix(``str``, n, k) ` ` `  `# This code is contributed by Surendra_Gangwar `

## C#

 `// C# implementation to find ` `// longest suffix of the String ` `// such that occurrence of each  ` `// character is less than K ` `using` `System; ` ` `  `class` `GFG{ ` `  `  `// Function to find the maximum ` `// length suffix in the String ` `static` `void` `maximumSuffix(String s,  ` `            ``int` `n, ``int` `k){ ` `      `  `    ``// Length of the String ` `    ``int` `i = s.Length - 1; ` `  `  `    ``// Map to store the number ` `    ``// of occurrence of character ` `    ``int` `[]arr = ``new` `int``; ` `    ``String suffix = ``""``; ` `      `  `    ``// Loop to iterate String  ` `    ``// from the last character ` `    ``while` `(i > -1) { ` `  `  `        ``int` `index = s[i] - ``'a'``; ` `          `  `        ``// Condition to check if the  ` `        ``// occurrence of each character ` `        ``// is less than given number ` `        ``if` `(arr[index] < n) { ` `            ``arr[index]++; ` `            ``suffix += s[i]; ` `            ``i--; ` `            ``continue``; ` `        ``} ` `          `  `        ``// Condition when character ` `        ``// cannot be deleted ` `        ``if` `(k == 0) ` `            ``break``; ` `        ``k--; ` `        ``i--; ` `    ``} ` `    ``suffix = reverse(suffix); ` `      `  `    ``// longest suffix ` `    ``Console.Write(suffix); ` `} ` `static` `String reverse(String input) { ` `    ``char``[] a = input.ToCharArray(); ` `    ``int` `l, r = a.Length - 1; ` `    ``for` `(l = 0; l < r; l++, r--) { ` `        ``char` `temp = a[l]; ` `        ``a[l] = a[r]; ` `        ``a[r] = temp; ` `    ``} ` `    ``return` `String.Join(``""``,a); ` `} ` `  `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String str = ``"iahagafedcba"``; ` `  `  `    ``int` `n = 1, k = 2; ` `      `  `    ``// Function call ` `    ``maximumSuffix(str, n, k); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```hgfedcba
```

Performance Analysis:

• Time Complexity: In the above-given approach, there is one loop for iterating over string which takes O(L) time in worst case. Therefore, the time complexity for this approach will be O(L).
• Auxiliary Space Complexity: In the above-given approach, there is extra space used to store the frequency of the character. Therefore, the auxiliary space complexity for the above approach will be O(L)

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