Lengths of maximized partitions of a string such that each character of the string appears in one substring

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++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
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;
}

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Java

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// 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<Integer> ans =
         new Vector<Integer>();
 
  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

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Python3

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# 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

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C#

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// 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

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Output: 

1 5





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

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