What is the time complexity of Minimum Substring Partition of Equal Character Frequency?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Minimum Substring Partition of Equal Character Frequency?

This problem uses Hash Table, String, Dynamic Programming, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Substring Partition of Equal Character Frequency?

Insight 1: Dynamic Programming can be used to efficiently solve this problem by storing the minimum number of substrings needed for partitions ending at each index. Insight 2: To determine if a substring is balanced, we need to count the frequency of each character and then check if all frequencies are equal. A `counts` array (size 26) and a `freq_counts` map (frequency of frequencies) are useful data structures.

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Minimum Substring Partition of Equal Character Frequency - LeetCode 3144 Solution

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Minimum Substring Partition of Equal Character Frequency - Complete Solution Guide

Minimum Substring Partition of Equal Character Frequency is LeetCode problem 3144, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s , you need to partition it into one or more balanced substrings . For example, if s == "ababcc" then ("abab", "c", "c") , ("ab", "abc", "c") , and ("ababcc") are all valid partitions, but ("a", "bab" , "cc") , ( "aba" , "bc", "c") , and ("ab", "abcc" ) are not. The unbalanced substrings are bolded. Return the minimum number of substrings that you can partition s into. Note: A balanced string is a string where each character in the string occurs the same number of times. Example

Detailed Explanation

The problem asks us to find the minimum number of substrings that a given string `s` can be partitioned into, such that each substring is 'balanced'. A 'balanced' string is defined as a string where each character in the string occurs the same number of times. For example, "aabb" is balanced because 'a' appears twice and 'b' appears twice. The input is a string `s` consisting of lowercase English letters, and the output is the minimum number of balanced substrings in a valid partition.

Solution Approach

The provided code uses dynamic programming to solve this problem. `dp[i]` stores the minimum number of substrings needed to partition the string up to index `i`. The algorithm iterates through each possible ending index `i` from 1 to n. For each `i`, it iterates backwards from `i-1` to 0 to find all possible starting positions `j` for a substring ending at `i`. It maintains a character count (`counts`) and a frequency count (`freq_counts`) for the substring `s[j:i]`. If the substring is balanced (i.e., all characters have the same frequency), the algorithm updates `dp[i]` with the minimum of its current value and `dp[j] + 1` (the minimum substrings to reach `j` plus one for the current substring).

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1`, where `dp[i]` stores the minimum number of balanced substrings to partition `s[0:i]`. Initially, `dp[i] = i` (worst case scenario, each character is a substring).
Step 2: Iterate from `i = 1` to `n`. This represents the ending index of the substring we're considering.
Step 3: For each `i`, iterate backwards from `j = i-1` to `0`. This represents the starting index of the substring ending at `i`.
Step 4: Maintain a `counts` array (size 26) to track the frequency of each character in the substring `s[j:i]`.
Step 5: Maintain a `freq_counts` map (or array for C) to track the frequency of the frequencies. For example, if `counts` is `[2, 2, 0, ..., 0]`, then `freq_counts` would be `{2: 2}` because the frequency `2` appears twice.
Step 6: Update the `counts` and `freq_counts` as we iterate backwards from `j = i-1` to `0`.
Step 7: Check if `freq_counts` has only one entry. If it does, this means that all characters in `s[j:i]` have the same frequency, making it a balanced string.
Step 8: If `s[j:i]` is balanced, update `dp[i] = min(dp[i], dp[j] + 1)`. This checks if using the balanced substring `s[j:i]` results in a smaller partition than the current minimum.
Step 9: After iterating through all possible `j` for a given `i`, `dp[i]` will contain the minimum number of balanced substrings needed to partition `s[0:i]`.
Step 10: Finally, return `dp[n]`, which represents the minimum number of balanced substrings needed to partition the entire string `s`.

Key Insights

Insight 1: Dynamic Programming can be used to efficiently solve this problem by storing the minimum number of substrings needed for partitions ending at each index.
Insight 2: To determine if a substring is balanced, we need to count the frequency of each character and then check if all frequencies are equal. A `counts` array (size 26) and a `freq_counts` map (frequency of frequencies) are useful data structures.
Insight 3: Optimization involves checking for balanced substrings in reverse order to find the optimal partitioning from the start of the string to any given index.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, String, Dynamic Programming, Counting.

Companies

Asked at: Mitsogo.

Minimum Substring Partition of Equal Character Frequency | LeetCode 3144

Medium

Hash Table String Dynamic Programming Counting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Arrays;

class Solution {
    public int minimumSubstringsInPartition(String s) {
        int n = s.length();
        int[] dp = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            dp[i] = i;
        }

        for (int i = 1; i <= n; i++) {
            int[] counts = new int[26];
            HashMap<Integer, Integer> freq_counts = new HashMap<>();
            for (int j = i - 1; j >= 0; j--) {
                int char_idx = s.charAt(j) - 'a';

                int old_freq = counts[char_idx];
                if (old_freq > 0) {
                    freq_counts.put(old_freq, freq_counts.get(old_freq) - 1);
                    if (freq_counts.get(old_freq) == 0) {
                        freq_counts.remove(old_freq);
                    }
                }

                counts[char_idx] += 1;
                int new_freq = counts[char_idx];
                freq_counts.put(new_freq, freq_counts.getOrDefault(new_freq, 0) + 1);

                if (freq_counts.size() == 1) {
                    if (j > 0) {
                        dp[i] = Math.min(dp[i], dp[j] + 1);
                    } else {
                        dp[i] = Math.min(dp[i], 1);
                    }
                }
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 3144: Minimum Substring Partition of Equal Character Frequency - Detailed explanation and solution approaches.

Topics

Hash Table String Dynamic Programming Counting

Companies

Mitsogo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3144

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a DP array `dp` of size `n+1`, where `dp[i]` stores the minimum number of balanced substrings to partition `s[0:i]`. Initially, `dp[i] = i` (worst case scenario, each character is a substring).
2
Step 2: Iterate from `i = 1` to `n`. This represents the ending index of the substring we're considering.
3
Step 3: For each `i`, iterate backwards from `j = i-1` to `0`. This represents the starting index of the substring ending at `i`.
4
Step 4: Maintain a `counts` array (size 26) to track the frequency of each character in the substring `s[j:i]`.
5
Step 5: Maintain a `freq_counts` map (or array for C) to track the frequency of the frequencies. For example, if `counts` is `[2, 2, 0, ..., 0]`, then `freq_counts` would be `{2: 2}` because the frequency `2` appears twice.
6
Step 6: Update the `counts` and `freq_counts` as we iterate backwards from `j = i-1` to `0`.
7
Step 7: Check if `freq_counts` has only one entry. If it does, this means that all characters in `s[j:i]` have the same frequency, making it a balanced string.
8
Step 8: If `s[j:i]` is balanced, update `dp[i] = min(dp[i], dp[j] + 1)`. This checks if using the balanced substring `s[j:i]` results in a smaller partition than the current minimum.
9
Step 9: After iterating through all possible `j` for a given `i`, `dp[i]` will contain the minimum number of balanced substrings needed to partition `s[0:i]`.
10
Step 10: Finally, return `dp[n]`, which represents the minimum number of balanced substrings needed to partition the entire string `s`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops dominate the time complexity, resulting in O(n^2). The `dp` array is the main contributor to the space complexity, resulting in O(n). The character frequency arrays and maps have constant size relative to the input string length.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Greedy Approach

A greedy approach might try to find the longest balanced substring starting from the beginning of the string and then repeat the process for the remaining string. However, this approach is not guaranteed to find the optimal solution.

Time: O(n^2) in the worst case, potentially O(n) if a good balanced substring can be found quickly at each step.Space: O(1) or O(n) depending on whether substring extraction creates new strings or views of the original

Trade-offs: The greedy approach may be faster in some cases, but it's not guaranteed to find the minimum number of substrings. The dynamic programming approach guarantees the optimal solution at the cost of higher time complexity.

✨

Key Insights

1Insight 1: Dynamic Programming can be used to efficiently solve this problem by storing the minimum number of substrings needed for partitions ending at each index.
2Insight 2: To determine if a substring is balanced, we need to count the frequency of each character and then check if all frequencies are equal. A `counts` array (size 26) and a `freq_counts` map (frequency of frequencies) are useful data structures.
3Insight 3: Optimization involves checking for balanced substrings in reverse order to find the optimal partitioning from the start of the string to any given index.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The code initializes `dp[0]` to 0, handling the case of an empty string implicitly because the loop starts from i = 1.
Edge case 2: String with only one character. The loop executes and identifies this as a balanced substring, correctly setting `dp[1]` to 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly checking if a substring is balanced. Forgetting to handle cases where frequencies are updated by removing frequencies that have been decreased from count. Using `freq_counts` incorrectly can lead to counting unbalanced strings as balanced.

✗Common mistake 2: Not initializing the DP array correctly. Incorrect initialization may lead to wrong results. It's crucial to initialize `dp[i] = i` to represent the worst-case scenario of having each character as a separate substring.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. Finding balanced substrings can potentially be used to identify repeating patterns in data for efficient compression.

🏦Real-world scenario 2: Bioinformatics. Analyzing DNA sequences for balanced regions could provide insights into genomic structures or regulatory elements.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Substring Partition of Equal Character Frequency"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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