LeetCode #3144 — MEDIUM

Minimum Substring Partition of Equal Character Frequency

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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 1:

Input: s = "fabccddg"

Output: 3

Explanation:

We can partition the string s into 3 substrings in one of the following ways: ("fab, "ccdd", "g"), or ("fabc", "cd", "dg").

Example 2:

Input: s = "abababaccddb"

Output: 2

Explanation:

We can partition the string s into 2 substrings like so: ("abab", "abaccddb").

Constraints:

1 <= s.length <= 1000
s consists only of English lowercase letters.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Dynamic Programming

Example 1

"fabccddg"

Example 2

"abababaccddb"

Core Insight

What unlocks the optimal approach

Let <code>dp[i]</code> be the minimum number of partitions for the prefix ending at index <code>i + 1</code>.
<code>dp[i]</code> can be calculated as the <code>min(dp[j])</code> over all <code>j</code> such that <code>j < i</code> and <code>word[j+1…i]</code> is valid.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3144: Minimum Substring Partition of Equal Character Frequency
class Solution {
    private int n;
    private char[] s;
    private Integer[] f;

    public int minimumSubstringsInPartition(String s) {
        n = s.length();
        f = new Integer[n];
        this.s = s.toCharArray();
        return dfs(0);
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int[] cnt = new int[26];
        Map<Integer, Integer> freq = new HashMap<>(26);
        int ans = n - i;
        for (int j = i; j < n; ++j) {
            int k = s[j] - 'a';
            if (cnt[k] > 0) {
                if (freq.merge(cnt[k], -1, Integer::sum) == 0) {
                    freq.remove(cnt[k]);
                }
            }
            ++cnt[k];
            freq.merge(cnt[k], 1, Integer::sum);
            if (freq.size() == 1) {
                ans = Math.min(ans, 1 + dfs(j + 1));
            }
        }
        return f[i] = ans;
    }
}

// Accepted solution for LeetCode #3144: Minimum Substring Partition of Equal Character Frequency
func minimumSubstringsInPartition(s string) int {
	n := len(s)
	f := make([]int, n)
	for i := range f {
		f[i] = -1
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		cnt := [26]int{}
		freq := map[int]int{}
		f[i] = n - i
		for j := i; j < n; j++ {
			k := int(s[j] - 'a')
			if cnt[k] > 0 {
				freq[cnt[k]]--
				if freq[cnt[k]] == 0 {
					delete(freq, cnt[k])
				}
			}
			cnt[k]++
			freq[cnt[k]]++
			if len(freq) == 1 {
				f[i] = min(f[i], 1+dfs(j+1))
			}
		}
		return f[i]
	}
	return dfs(0)
}

# Accepted solution for LeetCode #3144: Minimum Substring Partition of Equal Character Frequency
class Solution:
    def minimumSubstringsInPartition(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            cnt = defaultdict(int)
            freq = defaultdict(int)
            ans = n - i
            for j in range(i, n):
                if cnt[s[j]]:
                    freq[cnt[s[j]]] -= 1
                    if not freq[cnt[s[j]]]:
                        freq.pop(cnt[s[j]])
                cnt[s[j]] += 1
                freq[cnt[s[j]]] += 1
                if len(freq) == 1 and (t := 1 + dfs(j + 1)) < ans:
                    ans = t
            return ans

        n = len(s)
        return dfs(0)

// Accepted solution for LeetCode #3144: Minimum Substring Partition of Equal Character Frequency
/**
 * [3144] Minimum Substring Partition of Equal Character Frequency
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

impl Solution {
    pub fn minimum_substrings_in_partition(s: String) -> i32 {
        let s: Vec<char> = s.chars().collect();
        let n = s.len();

        let mut dp = vec![i32::MAX; n + 1];
        let mut char_count = HashMap::new();
        dp[0] = 0;

        for i in 1..=n {
            let mut max_count = 0;
            char_count.clear();

            for j in (1..=i).rev() {
                let entry = char_count.entry(s[j - 1]).or_insert(0);
                *entry += 1;
                max_count = max_count.max(*entry);

                if max_count * char_count.len() == i - j + 1 && dp[j - 1] != i32::MAX {
                    dp[i] = dp[i].min(dp[j - 1] + 1);
                }
            }
        }

        dp[n]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3144() {
        assert_eq!(
            3,
            Solution::minimum_substrings_in_partition("fabccddg".to_owned())
        );
        assert_eq!(
            2,
            Solution::minimum_substrings_in_partition("abababaccddb".to_owned())
        );
    }
}

// Accepted solution for LeetCode #3144: Minimum Substring Partition of Equal Character Frequency
function minimumSubstringsInPartition(s: string): number {
    const n = s.length;
    const f: number[] = Array(n).fill(-1);
    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i] !== -1) {
            return f[i];
        }
        const cnt: Map<number, number> = new Map();
        const freq: Map<number, number> = new Map();
        f[i] = n - i;
        for (let j = i; j < n; ++j) {
            const k = s.charCodeAt(j) - 97;
            if (freq.has(cnt.get(k)!)) {
                freq.set(cnt.get(k)!, freq.get(cnt.get(k)!)! - 1);
                if (freq.get(cnt.get(k)!) === 0) {
                    freq.delete(cnt.get(k)!);
                }
            }
            cnt.set(k, (cnt.get(k) || 0) + 1);
            freq.set(cnt.get(k)!, (freq.get(cnt.get(k)!) || 0) + 1);
            if (freq.size === 1) {
                f[i] = Math.min(f[i], 1 + dfs(j + 1));
            }
        }
        return f[i];
    };
    return dfs(0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n × |\Sigma|)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.