LeetCode #2745 — MEDIUM

Construct the Longest New String

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

The Problem

Problem Statement

You are given three integers x, y, and z.

You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring.

Return the maximum possible length of the new string.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: x = 2, y = 5, z = 1
Output: 12
Explanation: We can concatenate the strings "BB", "AA", "BB", "AA", "BB", and "AB" in that order. Then, our new string is "BBAABBAABBAB". 
That string has length 12, and we can show that it is impossible to construct a string of longer length.

Example 2:

Input: x = 3, y = 2, z = 2
Output: 14
Explanation: We can concatenate the strings "AB", "AB", "AA", "BB", "AA", "BB", and "AA" in that order. Then, our new string is "ABABAABBAABBAA". 
That string has length 14, and we can show that it is impossible to construct a string of longer length.

Constraints:

1 <= x, y, z <= 50

Step 01

Brute Force Baseline

Problem summary: You are given three integers x, y, and z. You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring. Return the maximum possible length of the new string. A substring is a contiguous non-empty sequence of characters within a string.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming · Greedy

Example 1

2
5
1

Example 2

3
2
2

Step 02

Core Insight

What unlocks the optimal approach

It can be proved that ALL “AB”s can be used in the optimal solution. (1) If the final string starts with 'A', we can put all unused “AB”s at the very beginning. (2) If the final string starts with 'B' (meaning) it starts with “BB”, we can put all unused “AB”s after the 2nd 'B'.
Using “AB” doesn’t increase the number of “AA”s or “BB”s we can use. If we put an “AB” after “BB”, then we still need to append “AA” as before, so it doesn’t change the state.
We only need to consider strings “AA” and “BB”; we can either use the pattern “AABBAABB…” or the pattern “BBAABBAA…”, depending on which one of x and y is larger.

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 #2745: Construct the Longest New String
class Solution {
    public int longestString(int x, int y, int z) {
        if (x < y) {
            return (x * 2 + z + 1) * 2;
        }
        if (x > y) {
            return (y * 2 + z + 1) * 2;
        }
        return (x + y + z) * 2;
    }
}

// Accepted solution for LeetCode #2745: Construct the Longest New String
func longestString(x int, y int, z int) int {
	if x < y {
		return (x*2 + z + 1) * 2
	}
	if x > y {
		return (y*2 + z + 1) * 2
	}
	return (x + y + z) * 2
}

# Accepted solution for LeetCode #2745: Construct the Longest New String
class Solution:
    def longestString(self, x: int, y: int, z: int) -> int:
        if x < y:
            return (x * 2 + z + 1) * 2
        if x > y:
            return (y * 2 + z + 1) * 2
        return (x + y + z) * 2

// Accepted solution for LeetCode #2745: Construct the Longest New 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 #2745: Construct the Longest New String
// class Solution {
//     public int longestString(int x, int y, int z) {
//         if (x < y) {
//             return (x * 2 + z + 1) * 2;
//         }
//         if (x > y) {
//             return (y * 2 + z + 1) * 2;
//         }
//         return (x + y + z) * 2;
//     }
// }

// Accepted solution for LeetCode #2745: Construct the Longest New String
function longestString(x: number, y: number, z: number): number {
    if (x < y) {
        return (x * 2 + z + 1) * 2;
    }
    if (x > y) {
        return (y * 2 + z + 1) * 2;
    }
    return (x + y + z) * 2;
}

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(1)

Space

O(1)

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.