LeetCode #2513 — MEDIUM

Minimize the Maximum of Two Arrays

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

Solve on LeetCode
The Problem

Problem Statement

We have two arrays arr1 and arr2 which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions:

  • arr1 contains uniqueCnt1 distinct positive integers, each of which is not divisible by divisor1.
  • arr2 contains uniqueCnt2 distinct positive integers, each of which is not divisible by divisor2.
  • No integer is present in both arr1 and arr2.

Given divisor1, divisor2, uniqueCnt1, and uniqueCnt2, return the minimum possible maximum integer that can be present in either array.

Example 1:

Input: divisor1 = 2, divisor2 = 7, uniqueCnt1 = 1, uniqueCnt2 = 3
Output: 4
Explanation: 
We can distribute the first 4 natural numbers into arr1 and arr2.
arr1 = [1] and arr2 = [2,3,4].
We can see that both arrays satisfy all the conditions.
Since the maximum value is 4, we return it.

Example 2:

Input: divisor1 = 3, divisor2 = 5, uniqueCnt1 = 2, uniqueCnt2 = 1
Output: 3
Explanation: 
Here arr1 = [1,2], and arr2 = [3] satisfy all conditions.
Since the maximum value is 3, we return it.

Example 3:

Input: divisor1 = 2, divisor2 = 4, uniqueCnt1 = 8, uniqueCnt2 = 2
Output: 15
Explanation: 
Here, the final possible arrays can be arr1 = [1,3,5,7,9,11,13,15], and arr2 = [2,6].
It can be shown that it is not possible to obtain a lower maximum satisfying all conditions.

Constraints:

  • 2 <= divisor1, divisor2 <= 105
  • 1 <= uniqueCnt1, uniqueCnt2 < 109
  • 2 <= uniqueCnt1 + uniqueCnt2 <= 109
Patterns Used
🔍Modified Binary Search→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: We have two arrays arr1 and arr2 which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions: arr1 contains uniqueCnt1 distinct positive integers, each of which is not divisible by divisor1. arr2 contains uniqueCnt2 distinct positive integers, each of which is not divisible by divisor2. No integer is present in both arr1 and arr2. Given divisor1, divisor2, uniqueCnt1, and uniqueCnt2, return the minimum possible maximum integer that can be present in either array.

Baseline thinking

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

Pattern signal: Math · Binary Search

Example 1

2
7
1
3

Example 2

3
5
2
1

Example 3

2
4
8
2
Step 02

Core Insight

What unlocks the optimal approach

  • Use binary search to find smallest maximum element.
  • Add numbers divisible by x in nums2 and vice versa.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2513: Minimize the Maximum of Two Arrays
class Solution {
    public int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
        long divisor = lcm(divisor1, divisor2);
        long left = 1, right = 10000000000L;
        while (left < right) {
            long mid = (left + right) >> 1;
            long cnt1 = mid / divisor1 * (divisor1 - 1) + mid % divisor1;
            long cnt2 = mid / divisor2 * (divisor2 - 1) + mid % divisor2;
            long cnt = mid / divisor * (divisor - 1) + mid % divisor;
            if (cnt1 >= uniqueCnt1 && cnt2 >= uniqueCnt2 && cnt >= uniqueCnt1 + uniqueCnt2) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return (int) left;
    }

    private long lcm(int a, int b) {
        return (long) a * b / gcd(a, b);
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#29Divide Two Integersmedium→#50Pow(x, n)medium→#69Sqrt(x)easy→#268Missing Numbereasy→#367Valid Perfect Squareeasy→