LeetCode #2430 — HARD

Maximum Deletions on a String

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s consisting of only lowercase English letters. In one operation, you can:

Delete the entire string s, or
Delete the first i letters of s if the first i letters of s are equal to the following i letters in s, for any i in the range 1 <= i <= s.length / 2.

For example, if s = "ababc", then in one operation, you could delete the first two letters of s to get "abc", since the first two letters of s and the following two letters of s are both equal to "ab".

Return the maximum number of operations needed to delete all of s.

Example 1:

Input: s = "abcabcdabc"
Output: 2
Explanation:
- Delete the first 3 letters ("abc") since the next 3 letters are equal. Now, s = "abcdabc".
- Delete all the letters.
We used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed.
Note that in the second operation we cannot delete "abc" again because the next occurrence of "abc" does not happen in the next 3 letters.

Example 2:

Input: s = "aaabaab"
Output: 4
Explanation:
- Delete the first letter ("a") since the next letter is equal. Now, s = "aabaab".
- Delete the first 3 letters ("aab") since the next 3 letters are equal. Now, s = "aab".
- Delete the first letter ("a") since the next letter is equal. Now, s = "ab".
- Delete all the letters.
We used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed.

Example 3:

Input: s = "aaaaa"
Output: 5
Explanation: In each operation, we can delete the first letter of s.

Constraints:

1 <= s.length <= 4000
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of only lowercase English letters. In one operation, you can: Delete the entire string s, or Delete the first i letters of s if the first i letters of s are equal to the following i letters in s, for any i in the range 1 <= i <= s.length / 2. For example, if s = "ababc", then in one operation, you could delete the first two letters of s to get "abc", since the first two letters of s and the following two letters of s are both equal to "ab". Return the maximum number of operations needed to delete all of s.

Baseline thinking

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

Pattern signal: Dynamic Programming · String Matching

Example 1

"abcabcdabc"

Example 2

"aaabaab"

Example 3

"aaaaa"

Core Insight

What unlocks the optimal approach

We can use dynamic programming to find the answer. Create a 0-indexed dp array where dp[i] represents the maximum number of moves needed to remove the first i + 1 letters from s.
What should we do if there is an i where it is impossible to remove the first i + 1 letters?
Use a sentinel value such as -1 to show that it is impossible.
How can we quickly determine if two substrings of s are equal? We can use hashing.

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

Largest constraint values

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 #2430: Maximum Deletions on a String
class Solution {
    private int n;
    private Integer[] f;
    private int[][] g;

    public int deleteString(String s) {
        n = s.length();
        f = new Integer[n];
        g = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    g[i][j] = g[i + 1][j + 1] + 1;
                }
            }
        }
        return dfs(0);
    }

    private int dfs(int i) {
        if (i == n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        f[i] = 1;
        for (int j = 1; j <= (n - i) / 2; ++j) {
            if (g[i][i + j] >= j) {
                f[i] = Math.max(f[i], 1 + dfs(i + j));
            }
        }
        return f[i];
    }
}

// Accepted solution for LeetCode #2430: Maximum Deletions on a String
func deleteString(s string) int {
	n := len(s)
	g := make([][]int, n+1)
	for i := range g {
		g[i] = make([]int, n+1)
	}
	for i := n - 1; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			if s[i] == s[j] {
				g[i][j] = g[i+1][j+1] + 1
			}
		}
	}
	f := make([]int, n)
	var dfs func(int) int
	dfs = func(i int) int {
		if i == n {
			return 0
		}
		if f[i] > 0 {
			return f[i]
		}
		f[i] = 1
		for j := 1; j <= (n-i)/2; j++ {
			if g[i][i+j] >= j {
				f[i] = max(f[i], dfs(i+j)+1)
			}
		}
		return f[i]
	}
	return dfs(0)
}

# Accepted solution for LeetCode #2430: Maximum Deletions on a String
class Solution:
    def deleteString(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i == n:
                return 0
            ans = 1
            for j in range(1, (n - i) // 2 + 1):
                if s[i : i + j] == s[i + j : i + j + j]:
                    ans = max(ans, 1 + dfs(i + j))
            return ans

        n = len(s)
        return dfs(0)

// Accepted solution for LeetCode #2430: Maximum Deletions on a String
// 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 #2430: Maximum Deletions on a String
// class Solution {
//     private int n;
//     private Integer[] f;
//     private int[][] g;
// 
//     public int deleteString(String s) {
//         n = s.length();
//         f = new Integer[n];
//         g = new int[n + 1][n + 1];
//         for (int i = n - 1; i >= 0; --i) {
//             for (int j = i + 1; j < n; ++j) {
//                 if (s.charAt(i) == s.charAt(j)) {
//                     g[i][j] = g[i + 1][j + 1] + 1;
//                 }
//             }
//         }
//         return dfs(0);
//     }
// 
//     private int dfs(int i) {
//         if (i == n) {
//             return 0;
//         }
//         if (f[i] != null) {
//             return f[i];
//         }
//         f[i] = 1;
//         for (int j = 1; j <= (n - i) / 2; ++j) {
//             if (g[i][i + j] >= j) {
//                 f[i] = Math.max(f[i], 1 + dfs(i + j));
//             }
//         }
//         return f[i];
//     }
// }

// Accepted solution for LeetCode #2430: Maximum Deletions on a String
function deleteString(s: string): number {
    const n = s.length;
    const f: number[] = new Array(n).fill(0);
    const dfs = (i: number): number => {
        if (i == n) {
            return 0;
        }
        if (f[i] > 0) {
            return f[i];
        }
        f[i] = 1;
        for (let j = 1; j <= (n - i) >> 1; ++j) {
            if (s.slice(i, i + j) == s.slice(i + j, i + j + j)) {
                f[i] = Math.max(f[i], dfs(i + 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^2)

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.

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.