LeetCode #1525 — MEDIUM

Number of Good Ways to Split a String

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

The Problem

Problem Statement

You are given a string s.

A split is called good if you can split s into two non-empty strings s_left and s_right where their concatenation is equal to s (i.e., s_left + s_right = s) and the number of distinct letters in s_left and s_right is the same.

Return the number of good splits you can make in s.

Example 1:

Input: s = "aacaba"
Output: 2
Explanation: There are 5 ways to split "aacaba" and 2 of them are good. 
("a", "acaba") Left string and right string contains 1 and 3 different letters respectively.
("aa", "caba") Left string and right string contains 1 and 3 different letters respectively.
("aac", "aba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aaca", "ba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aacab", "a") Left string and right string contains 3 and 1 different letters respectively.

Example 2:

Input: s = "abcd"
Output: 1
Explanation: Split the string as follows ("ab", "cd").

Constraints:

1 <= s.length <= 10⁵
s consists of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s. A split is called good if you can split s into two non-empty strings sleft and sright where their concatenation is equal to s (i.e., sleft + sright = s) and the number of distinct letters in sleft and sright is the same. Return the number of good splits you can make in s.

Baseline thinking

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

Pattern signal: Hash Map · Dynamic Programming · Bit Manipulation

Example 1

"aacaba"

Example 2

"abcd"

Step 02

Core Insight

What unlocks the optimal approach

Use two HashMap to store the counts of distinct letters in the left and right substring divided by the current index.

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 #1525: Number of Good Ways to Split a String
class Solution {
    public int numSplits(String s) {
        Map<Character, Integer> cnt = new HashMap<>();
        for (char c : s.toCharArray()) {
            cnt.merge(c, 1, Integer::sum);
        }
        Set<Character> vis = new HashSet<>();
        int ans = 0;
        for (char c : s.toCharArray()) {
            vis.add(c);
            if (cnt.merge(c, -1, Integer::sum) == 0) {
                cnt.remove(c);
            }
            if (vis.size() == cnt.size()) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1525: Number of Good Ways to Split a String
func numSplits(s string) (ans int) {
	cnt := map[rune]int{}
	for _, c := range s {
		cnt[c]++
	}
	vis := map[rune]bool{}
	for _, c := range s {
		vis[c] = true
		cnt[c]--
		if cnt[c] == 0 {
			delete(cnt, c)
		}
		if len(vis) == len(cnt) {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1525: Number of Good Ways to Split a String
class Solution:
    def numSplits(self, s: str) -> int:
        cnt = Counter(s)
        vis = set()
        ans = 0
        for c in s:
            vis.add(c)
            cnt[c] -= 1
            if cnt[c] == 0:
                cnt.pop(c)
            ans += len(vis) == len(cnt)
        return ans

// Accepted solution for LeetCode #1525: Number of Good Ways to Split a String
struct Solution;

impl Solution {
    fn num_splits(s: String) -> i32 {
        let mut total = vec![0; 26];
        for b in s.bytes() {
            total[(b - b'a') as usize] += 1;
        }

        let mut left = vec![0; 26];
        let mut right = total.to_vec();
        let mut res = 0;
        for b in s.bytes() {
            left[(b - b'a') as usize] += 1;
            right[(b - b'a') as usize] -= 1;
            let mut diff = 0;
            for i in 0..26 {
                if left[i] > 0 {
                    diff += 1;
                }
                if right[i] > 0 {
                    diff -= 1;
                }
            }
            if diff == 0 {
                res += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "aacaba".to_string();
    let res = 2;
    assert_eq!(Solution::num_splits(s), res);
    let s = "abcd".to_string();
    let res = 1;
    assert_eq!(Solution::num_splits(s), res);
    let s = "aaaaa".to_string();
    let res = 4;
    assert_eq!(Solution::num_splits(s), res);
    let s = "acbadbaada".to_string();
    let res = 2;
    assert_eq!(Solution::num_splits(s), res);
}

// Accepted solution for LeetCode #1525: Number of Good Ways to Split a String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1525: Number of Good Ways to Split a String
// class Solution {
//     public int numSplits(String s) {
//         Map<Character, Integer> cnt = new HashMap<>();
//         for (char c : s.toCharArray()) {
//             cnt.merge(c, 1, Integer::sum);
//         }
//         Set<Character> vis = new HashSet<>();
//         int ans = 0;
//         for (char c : s.toCharArray()) {
//             vis.add(c);
//             if (cnt.merge(c, -1, Integer::sum) == 0) {
//                 cnt.remove(c);
//             }
//             if (vis.size() == cnt.size()) {
//                 ++ans;
//             }
//         }
//         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.