LeetCode #3557 — MEDIUM

Find Maximum Number of Non Intersecting Substrings

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

The Problem

Problem Statement

You are given a string word.

Return the maximum number of non-intersecting substrings of word that are at least four characters long and start and end with the same letter.

Example 1:

Input: word = "abcdeafdef"

Output: 2

Explanation:

The two substrings are "abcdea" and "fdef".

Example 2:

Input: word = "bcdaaaab"

Output: 1

Explanation:

The only substring is "aaaa". Note that we cannot also choose "bcdaaaab" since it intersects with the other substring.

Constraints:

1 <= word.length <= 2 * 10⁵
word consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string word. Return the maximum number of non-intersecting substrings of word that are at least four characters long and start and end with the same letter.

Baseline thinking

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

Pattern signal: Hash Map · Dynamic Programming · Greedy

Example 1

"abcdeafdef"

Example 2

"bcdaaaab"

Step 02

Core Insight

What unlocks the optimal approach

Can we solve the problem using Dynamic Programming?
For each character <code>c</code>, store all occurrence indices in order
At each position <code>i</code>, let <code>j</code> be the first index of <code>word[i]</code>; if <code>i - j >= 3</code>, we can form substring <code>[j, i]</code>
For each index, also store the maximum for <b>any</b> substring ending before that index in the dp.

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 #3557: Find Maximum Number of Non Intersecting Substrings
class Solution {
  public int maxSubstrings(String word) {
    int ans = 0;
    Map<Character, Integer> firstSeen = new HashMap<>();

    for (int i = 0; i < word.length(); ++i) {
      final char c = word.charAt(i);
      if (!firstSeen.containsKey(c)) {
        firstSeen.put(c, i);
      } else if (i - firstSeen.get(c) + 1 >= 4) {
        ++ans;
        firstSeen.clear();
      }
    }

    return ans;
  }
}

// Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
package main

// https://space.bilibili.com/206214
func maxSubstrings1(word string) (ans int) {
	pos := [26]int{}
	for i, b := range word {
		b -= 'a'
		if pos[b] == 0 {
			pos[b] = i + 1
		} else if i-pos[b] > 1 {
			ans++
			clear(pos[:])
		}
	}
	return
}

func maxSubstrings(word string) (ans int) {
	seen := 0
	for i := 3; i < len(word); i++ {
		seen |= 1 << (word[i-3] - 'a')
		if seen>>(word[i]-'a')&1 > 0 { // 再次遇到 word[i]
			ans++
			seen = 0
			i += 3
		}
	}
	return
}

func maxSubstringsDP(word string) int {
	pos := [26][]int{}
	n := len(word)
	f := make([]int, n+1)
	for i, b := range word {
		b -= 'a'
		for len(pos[b]) > 1 && i-pos[b][1] > 2 {
			pos[b] = pos[b][1:]
		}
		f[i+1] = f[i] // 不选 s[i]
		if len(pos[b]) > 0 && i-pos[b][0] > 2 {
			f[i+1] = max(f[i+1], f[pos[b][0]]+1) // 选 s[i]
		}
		pos[b] = append(pos[b], i)
	}
	return f[n]
}

# Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
class Solution:
  def maxSubstrings(self, word: str) -> int:
    ans = 0
    firstSeen = {}

    for i, c in enumerate(word):
      if c not in firstSeen:
        firstSeen[c] = i
      elif i - firstSeen[c] + 1 >= 4:
        ans += 1
        firstSeen.clear()

    return ans

// Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
// class Solution {
//   public int maxSubstrings(String word) {
//     int ans = 0;
//     Map<Character, Integer> firstSeen = new HashMap<>();
// 
//     for (int i = 0; i < word.length(); ++i) {
//       final char c = word.charAt(i);
//       if (!firstSeen.containsKey(c)) {
//         firstSeen.put(c, i);
//       } else if (i - firstSeen.get(c) + 1 >= 4) {
//         ++ans;
//         firstSeen.clear();
//       }
//     }
// 
//     return ans;
//   }
// }

// Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3557: Find Maximum Number of Non Intersecting Substrings
// class Solution {
//   public int maxSubstrings(String word) {
//     int ans = 0;
//     Map<Character, Integer> firstSeen = new HashMap<>();
// 
//     for (int i = 0; i < word.length(); ++i) {
//       final char c = word.charAt(i);
//       if (!firstSeen.containsKey(c)) {
//         firstSeen.put(c, i);
//       } else if (i - firstSeen.get(c) + 1 >= 4) {
//         ++ans;
//         firstSeen.clear();
//       }
//     }
// 
//     return ans;
//   }
// }

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 × m)

Space

O(n × m)

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.