Skip to content
Related Articles
Longest suffix such that occurrence of each character is less than N after deleting atmost K characters
• Last Updated : 29 May, 2021

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.

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

• Initialize 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)

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 industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up