Construct the Longest New String - Practice Coding Problems

Construct the Longest New String - Problem

Imagine you're working with a string construction system where you have three types of building blocks:

x blocks of type "AA"
y blocks of type "BB"
z blocks of type "AB"

Your goal is to concatenate some or all of these blocks in any order to create the longest possible string with one critical constraint: the final string must not contain "AAA" or "BBB" as a substring.

For example, if you have x=2, y=1, z=1, you have blocks ["AA", "AA", "BB", "AB"]. You could arrange them as "AA" + "BB" + "AA" + "AB" = "AABBAAB" (length 7), but you need to ensure no three consecutive A's or B's appear.

Return the maximum possible length of such a valid string.

Input & Output

example_1.py — Basic Case

$ Input: x = 2, y = 5, z = 1

› Output: 16

💡 Note: We have 2 "AA", 5 "BB", and 1 "AB" blocks. Since both x > 0 and y > 0, we can use all blocks. Total length = (2 + 5 + 1) × 2 = 16. One valid arrangement: "AA" + "BB" + "AA" + "AB" + "BB" + "BB" + "BB" + "BB".

example_2.py — No AA Blocks

$ Input: x = 0, y = 3, z = 1

› Output: 4

💡 Note: We have 0 "AA", 3 "BB", and 1 "AB" blocks. Since x = 0, we can use at most 1 "BB" block. Result = min(1, 3) × 2 + 1 × 2 = 2 + 2 = 4. Valid arrangement: "BB" + "AB".

example_3.py — No BB Blocks

$ Input: x = 3, y = 0, z = 1

› Output: 4

💡 Note: We have 3 "AA", 0 "BB", and 1 "AB" blocks. Since y = 0, we can use at most 1 "AA" block. Result = min(1, 3) × 2 + 1 × 2 = 2 + 2 = 4. Valid arrangement: "AA" + "AB".

Visualization

Tap to expand

Understanding the Visualization

Identify Block Types

AA blocks (red), BB blocks (blue), AB blocks (mixed) each contribute 2 characters

Recognize Constraints

Cannot have AAA or BBB sequences in the final string

Use AB as Separators

AB blocks can break up potentially problematic AA-AA or BB-BB sequences

Apply Mathematical Formula

Calculate directly based on whether both AA and BB blocks are available

Key Takeaway

🎯 Key Insight: AB blocks are the most valuable because they act as separators, preventing forbidden AAA or BBB patterns while contributing 2 characters each. Mathematical analysis eliminates the need for expensive permutation testing.

Time & Space Complexity

Time Complexity

⏱️

O(1)

Just a few mathematical calculations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only uses a constant amount of extra variables

✓ Linear Space

Constraints

0 ≤ x, y, z ≤ 50
At least one of x, y, z is positive
Total blocks will not exceed 150

Asked in

G Google 25 f Meta 18 a Amazon 15 ⊞ Microsoft 12

The optimal solution uses mathematical analysis instead of brute force. The key insight is that AB blocks act as separators that prevent AAA and BBB patterns. When both x > 0 and y > 0, we can use all blocks by alternating patterns. When one is zero, we're limited to at most one block of the other type. Time complexity: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Analysis (Optimal)	O(1)	O(1)	Use mathematical insight to calculate maximum length directly

Mathematical Analysis (Optimal) — Algorithm Steps

Recognize that each block contributes characters: AA→2A, BB→2B, AB→1A+1B
Total A's = 2x + z, Total B's = 2y + z
AB blocks are most valuable as they can separate AA and BB blocks
Calculate maximum usable blocks based on the constraint that we can't have 3 consecutive same characters
Use the insight that we can always use all AB blocks, and the limiting factor is the balance between A's and B's

Visualization

Tap to expand

Step-by-Step Walkthrough

Analyze Block Types

AA and BB are restrictive, AB is flexible

Find Pattern

AB blocks can separate AA and BB blocks effectively

Handle Edge Cases

When x=0 or y=0, we're more limited

Calculate Directly

Use mathematical formula instead of trying combinations

Code -

solution.c — C

#include <stdio.h>

int min(int a, int b) {
    return a < b ? a : b;
}

int longestString(int x, int y, int z) {
    if (x == 0) {
        return min(1, y) * 2 + z * 2;
    }
    if (y == 0) {
        return min(1, x) * 2 + z * 2;
    }
    
    return (x + y + z) * 2;
}

int main() {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    printf("%d\n", longestString(x, y, z));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Just a few mathematical calculations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only uses a constant amount of extra variables

✓ Linear Space

Constraints

0 ≤ x, y, z ≤ 50
At least one of x, y, z is positive
Total blocks will not exceed 150

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Mathematical Analysis (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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