LeetCode #2468 — HARD

Split Message Based on Limit

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

The Problem

Problem Statement

You are given a string, message, and a positive integer, limit.

You must split message into one or more parts based on limit. Each resulting part should have the suffix "<a/b>", where "b" is to be replaced with the total number of parts and "a" is to be replaced with the index of the part, starting from 1 and going up to b. Additionally, the length of each resulting part (including its suffix) should be equal to limit, except for the last part whose length can be at most limit.

The resulting parts should be formed such that when their suffixes are removed and they are all concatenated in order, they should be equal to message. Also, the result should contain as few parts as possible.

Return the parts message would be split into as an array of strings. If it is impossible to split message as required, return an empty array.

Example 1:

Input: message = "this is really a very awesome message", limit = 9
Output: ["thi<1/14>","s i<2/14>","s r<3/14>","eal<4/14>","ly <5/14>","a v<6/14>","ery<7/14>"," aw<8/14>","eso<9/14>","me<10/14>"," m<11/14>","es<12/14>","sa<13/14>","ge<14/14>"]
Explanation:
The first 9 parts take 3 characters each from the beginning of message.
The next 5 parts take 2 characters each to finish splitting message. 
In this example, each part, including the last, has length 9. 
It can be shown it is not possible to split message into less than 14 parts.

Example 2:

Input: message = "short message", limit = 15
Output: ["short mess<1/2>","age<2/2>"]
Explanation:
Under the given constraints, the string can be split into two parts: 
- The first part comprises of the first 10 characters, and has a length 15.
- The next part comprises of the last 3 characters, and has a length 8.

Constraints:

1 <= message.length <= 10⁴
message consists only of lowercase English letters and ' '.
1 <= limit <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given a string, message, and a positive integer, limit. You must split message into one or more parts based on limit. Each resulting part should have the suffix "<a/b>", where "b" is to be replaced with the total number of parts and "a" is to be replaced with the index of the part, starting from 1 and going up to b. Additionally, the length of each resulting part (including its suffix) should be equal to limit, except for the last part whose length can be at most limit. The resulting parts should be formed such that when their suffixes are removed and they are all concatenated in order, they should be equal to message. Also, the result should contain as few parts as possible. Return the parts message would be split into as an array of strings. If it is impossible to split message as required, return an empty array.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"this is really a very awesome message"
9

Example 2

"short message"
15

Core Insight

What unlocks the optimal approach

Could you solve the problem if you knew how many digits the total number of parts has?
Try all possible lengths of the total number of parts, and see if the string can be split such that the total number of parts has that length.
Binary search can be used for each part length to find the precise number of parts needed.

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 #2468: Split Message Based on Limit
class Solution {
    public String[] splitMessage(String message, int limit) {
        int n = message.length();
        int sa = 0;
        String[] ans = new String[0];
        for (int k = 1; k <= n; ++k) {
            int lk = (k + "").length();
            sa += lk;
            int sb = lk * k;
            int sc = 3 * k;
            if (limit * k - (sa + sb + sc) >= n) {
                int i = 0;
                ans = new String[k];
                for (int j = 1; j <= k; ++j) {
                    String tail = String.format("<%d/%d>", j, k);
                    String t = message.substring(i, Math.min(n, i + limit - tail.length())) + tail;
                    ans[j - 1] = t;
                    i += limit - tail.length();
                }
                break;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2468: Split Message Based on Limit
func splitMessage(message string, limit int) (ans []string) {
	n := len(message)
	sa := 0
	for k := 1; k <= n; k++ {
		lk := len(strconv.Itoa(k))
		sa += lk
		sb := lk * k
		sc := 3 * k
		if limit*k-(sa+sb+sc) >= n {
			i := 0
			for j := 1; j <= k; j++ {
				tail := "<" + strconv.Itoa(j) + "/" + strconv.Itoa(k) + ">"
				t := message[i:min(i+limit-len(tail), n)] + tail
				ans = append(ans, t)
				i += limit - len(tail)
			}
			break
		}
	}
	return
}

# Accepted solution for LeetCode #2468: Split Message Based on Limit
class Solution:
    def splitMessage(self, message: str, limit: int) -> List[str]:
        n = len(message)
        sa = 0
        for k in range(1, n + 1):
            sa += len(str(k))
            sb = len(str(k)) * k
            sc = 3 * k
            if limit * k - (sa + sb + sc) >= n:
                ans = []
                i = 0
                for j in range(1, k + 1):
                    tail = f'<{j}/{k}>'
                    t = message[i : i + limit - len(tail)] + tail
                    ans.append(t)
                    i += limit - len(tail)
                return ans
        return []

// Accepted solution for LeetCode #2468: Split Message Based on Limit
// 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 #2468: Split Message Based on Limit
// class Solution {
//     public String[] splitMessage(String message, int limit) {
//         int n = message.length();
//         int sa = 0;
//         String[] ans = new String[0];
//         for (int k = 1; k <= n; ++k) {
//             int lk = (k + "").length();
//             sa += lk;
//             int sb = lk * k;
//             int sc = 3 * k;
//             if (limit * k - (sa + sb + sc) >= n) {
//                 int i = 0;
//                 ans = new String[k];
//                 for (int j = 1; j <= k; ++j) {
//                     String tail = String.format("<%d/%d>", j, k);
//                     String t = message.substring(i, Math.min(n, i + limit - tail.length())) + tail;
//                     ans[j - 1] = t;
//                     i += limit - tail.length();
//                 }
//                 break;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2468: Split Message Based on Limit
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2468: Split Message Based on Limit
// class Solution {
//     public String[] splitMessage(String message, int limit) {
//         int n = message.length();
//         int sa = 0;
//         String[] ans = new String[0];
//         for (int k = 1; k <= n; ++k) {
//             int lk = (k + "").length();
//             sa += lk;
//             int sb = lk * k;
//             int sc = 3 * k;
//             if (limit * k - (sa + sb + sc) >= n) {
//                 int i = 0;
//                 ans = new String[k];
//                 for (int j = 1; j <= k; ++j) {
//                     String tail = String.format("<%d/%d>", j, k);
//                     String t = message.substring(i, Math.min(n, i + limit - tail.length())) + tail;
//                     ans[j - 1] = t;
//                     i += limit - tail.length();
//                 }
//                 break;
//             }
//         }
//         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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.