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Lengths of maximized partitions of a string such that each character of the string appears in one substring
• Difficulty Level : Medium
• Last Updated : 26 Sep, 2020

Given string str of lowercase alphabets, split the given string into as many substrings as possible such that each character from the given string appears in a single substring. The task is to print the length of all such partitions.

Examples:

Input: str = “acbbcc”
Output: 1 5
Explanation:
Possible partitions where each character of the strings occurs in at most one partition are “a” and “cbbcc”.
Therefore, the length is {1, 5}

Input: str = “abaccbdeffed”
Output: 6 6
Explanation:
Possible partitions where each character of the strings occurs in at most one partition are “abaccb” and “deffed”.
Therefore, the length is {6, 6}

Approach: This problem can be solved easily by using the Greedy Approach. Follow the steps given below to solve the problem.

1. Store the last index of all characters in the string.
2. Since the string contains only lowercase letters, simply use an array of fixed size 26 to store the last indices of each character.
3. Iterate over the given string and follow the steps below:
• Add the current character to the partition if the last position of the character exceeds the current index and increase the length of the partition.
• If the last index of the current character is equal to the current index, then store its length and proceed to the next character and so on.
4. After completing the above steps, print all the lengths stored.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the length of` `// all partitions oof a string such` `// that each characters occurs in` `// a single substring` `void` `partitionString(string s)` `{` `    ``int` `n = s.size();`   `    ``// Stores last index of string s` `    ``vector<``int``> ans;`   `    ``if` `(n == 0) {` `        ``cout << ``"-1"``;` `        ``return``;` `    ``}`   `    ``// Find the last position of` `    ``// each letter in the string` `    ``vector<``int``> last_pos(26, -1);`   `    ``for` `(``int` `i = n - 1; i >= 0; --i) {`   `        ``// Update the last index` `        ``if` `(last_pos[s[i] - ``'a'``] == -1) {` `            ``last_pos[s[i] - ``'a'``] = i;` `        ``}` `    ``}`   `    ``int` `minp = -1, plen = 0;`   `    ``// Iterate the given string` `    ``for` `(``int` `i = 0; i < n; ++i) {`   `        ``// Get the last index of s[i]` `        ``int` `lp = last_pos[s[i] - ``'a'``];`   `        ``// Extend the current partition` `        ``// characters last pos` `        ``minp = max(minp, lp);`   `        ``// Increase len of partition` `        ``++plen;`   `        ``// if the current pos of` `        ``// character equals the min pos` `        ``// then the end of partition` `        ``if` `(i == minp) {`   `            ``// Store the length` `            ``ans.push_back(plen);` `            ``minp = -1;` `            ``plen = 0;` `        ``}` `    ``}`   `    ``// Print all the partition lengths` `    ``for` `(``int` `i = 0;` `         ``i < (``int``)ans.size(); i++) {` `        ``cout << ans[i] << ``" "``;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given string str` `    ``string str = ``"acbbcc"``;`   `    ``// Function Call` `    ``partitionString(str);`   `    ``return` `0;` `}`

## Java

 `// Java program for ` `// the above approach` `import` `java.util.*;` `class` `GFG{`   `// Function to find the length of` `// all partitions oof a String such` `// that each characters occurs in` `// a single subString` `static` `void` `partitionString(String s)` `{` `  ``int` `n = s.length();`   `  ``// Stores last index of String s` `  ``Vector ans = ` `         ``new` `Vector();`   `  ``if` `(n == ``0``) ` `  ``{` `    ``System.out.print(``"-1"``);` `    ``return``;` `  ``}`   `  ``// Find the last position of` `  ``// each letter in the String` `  ``int` `[]last_pos = ``new` `int``[``26``];` `  ``Arrays.fill(last_pos, -``1``);`   `  ``for` `(``int` `i = n - ``1``; i >= ``0``; --i) ` `  ``{` `    ``// Update the last index` `    ``if` `(last_pos[s.charAt(i) - ``'a'``] == -``1``) ` `    ``{` `      ``last_pos[s.charAt(i) - ``'a'``] = i;` `    ``}` `  ``}`   `  ``int` `minp = -``1``, plen = ``0``;`   `  ``// Iterate the given String` `  ``for` `(``int` `i = ``0``; i < n; ++i) ` `  ``{` `    ``// Get the last index of s[i]` `    ``int` `lp = last_pos[s.charAt(i) - ``'a'``];`   `    ``// Extend the current partition` `    ``// characters last pos` `    ``minp = Math.max(minp, lp);`   `    ``// Increase len of partition` `    ``++plen;`   `    ``// if the current pos of` `    ``// character equals the min pos` `    ``// then the end of partition` `    ``if` `(i == minp) ` `    ``{` `      ``// Store the length` `      ``ans.add(plen);` `      ``minp = -``1``;` `      ``plen = ``0``;` `    ``}` `  ``}`   `  ``// Print all the partition lengths` `  ``for` `(``int` `i = ``0``; i < (``int``)ans.size(); i++) ` `  ``{` `    ``System.out.print(ans.get(i) + ``" "``);` `  ``}` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `  ``// Given String str` `  ``String str = ``"acbbcc"``;`   `  ``// Function Call` `  ``partitionString(str);` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program for the above approach`   `# Function to find the length of` `# all partitions oof a string such` `# that each characters occurs in` `# a single substring` `def` `partitionString(s):` `    `  `    ``n ``=` `len``(s)`   `    ``# Stores last index of string s` `    ``ans ``=` `[]`   `    ``if` `(n ``=``=` `0``):` `        ``print``(``"-1"``)` `        ``return` `    `  `    ``# Find the last position of` `    ``# each letter in the string` `    ``last_pos ``=` `[``-``1``] ``*` `26`   `    ``for` `i ``in` `range``(n ``-` `1` `, ``-``1` `, ``-``1``):`   `        ``# Update the last index` `        ``if` `(last_pos[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=``=` `-``1``):` `            ``last_pos[``ord``(s[i]) ``-` `ord``(``'a'``)] ``=` `i` `    `  `    ``minp ``=` `-``1` `    ``plen ``=` `0`   `    ``# Iterate the given string` `    ``for` `i ``in` `range``(n):`   `        ``# Get the last index of s[i]` `        ``lp ``=` `last_pos[``ord``(s[i]) ``-` `ord``(``'a'``)]`   `        ``# Extend the current partition` `        ``# characters last pos` `        ``minp ``=` `max``(minp, lp)`   `        ``# Increase len of partition` `        ``plen ``+``=` `1`   `        ``# if the current pos of` `        ``# character equals the min pos` `        ``# then the end of partition` `        ``if` `(i ``=``=` `minp):`   `            ``# Store the length` `            ``ans.append(plen)` `            ``minp ``=` `-``1` `            ``plen ``=` `0` `        `  `    ``# Print all the partition lengths` `    ``for` `i ``in` `range``(``len``(ans)):` `        ``print``(ans[i], end ``=` `" "``)`   `# Driver Code`   `# Given string str` `str` `=` `"acbbcc"`   `# Function call` `partitionString(``str``)`   `# This code is contributed by code_hunt`

## C#

 `// C# program for ` `// the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{`   `// Function to find the length of` `// all partitions oof a String such` `// that each characters occurs in` `// a single subString` `static` `void` `partitionString(String s)` `{` `  ``int` `n = s.Length;`   `  ``// Stores last index of String s` `  ``List<``int``> ans = ``new` `List<``int``>();`   `  ``if` `(n == 0) ` `  ``{` `    ``Console.Write(``"-1"``);` `    ``return``;` `  ``}`   `  ``// Find the last position of` `  ``// each letter in the String` `  ``int` `[]last_pos = ``new` `int``[26];` `  ``for` `(``int` `i = 0; i < 26; ++i)` `  ``{` `    ``last_pos[i] = -1;  ` `  ``}`   `  ``for` `(``int` `i = n - 1; i >= 0; --i) ` `  ``{` `    ``// Update the last index` `    ``if` `(last_pos[s[i] - ``'a'``] == -1) ` `    ``{` `      ``last_pos[s[i] - ``'a'``] = i;` `    ``}` `  ``}`   `  ``int` `minp = -1, plen = 0;`   `  ``// Iterate the given String` `  ``for` `(``int` `i = 0; i < n; ++i) ` `  ``{` `    ``// Get the last index of s[i]` `    ``int` `lp = last_pos[s[i] - ``'a'``];`   `    ``// Extend the current partition` `    ``// characters last pos` `    ``minp = Math.Max(minp, lp);`   `    ``// Increase len of partition` `    ``++plen;`   `    ``// if the current pos of` `    ``// character equals the min pos` `    ``// then the end of partition` `    ``if` `(i == minp) ` `    ``{` `      ``// Store the length` `      ``ans.Add(plen);` `      ``minp = -1;` `      ``plen = 0;` `    ``}` `  ``}`   `  ``// Print all the partition lengths` `  ``for` `(``int` `i = 0; i < (``int``)ans.Count; i++) ` `  ``{` `    ``Console.Write(ans[i] + ``" "``);` `  ``}` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `  ``// Given String str` `  ``String str = ``"acbbcc"``;`   `  ``// Function Call` `  ``partitionString(str);` `}` `}`   `// This code is contributed by Rajput-Ji`

Output:

```1 5

```

Time Complexity: O(N)
Auxiliary Space: O(N)

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