LeetCode #321 — HARD

Create Maximum Number

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

Solve on LeetCode

The Problem

Problem Statement

You are given two integer arrays nums1 and nums2 of lengths m and n respectively. nums1 and nums2 represent the digits of two numbers. You are also given an integer k.

Create the maximum number of length k <= m + n from digits of the two numbers. The relative order of the digits from the same array must be preserved.

Return an array of the k digits representing the answer.

Example 1:

Input: nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], k = 5
Output: [9,8,6,5,3]

Example 2:

Input: nums1 = [6,7], nums2 = [6,0,4], k = 5
Output: [6,7,6,0,4]

Example 3:

Input: nums1 = [3,9], nums2 = [8,9], k = 3
Output: [9,8,9]

Constraints:

m == nums1.length
n == nums2.length
1 <= m, n <= 500
0 <= nums1[i], nums2[i] <= 9
1 <= k <= m + n
nums1 and nums2 do not have leading zeros.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays nums1 and nums2 of lengths m and n respectively. nums1 and nums2 represent the digits of two numbers. You are also given an integer k. Create the maximum number of length k <= m + n from digits of the two numbers. The relative order of the digits from the same array must be preserved. Return an array of the k digits representing the answer.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Stack · Greedy

Example 1

[3,4,6,5]
[9,1,2,5,8,3]
5

Example 2

[6,7]
[6,0,4]
5

Example 3

[3,9]
[8,9]
3

Related Problems

Remove K Digits (remove-k-digits)
Maximum Swap (maximum-swap)

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #321: Create Maximum Number
class Solution {
    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
        int m = nums1.length, n = nums2.length;
        int l = Math.max(0, k - n), r = Math.min(k, m);
        int[] ans = new int[k];
        for (int x = l; x <= r; ++x) {
            int[] arr1 = f(nums1, x);
            int[] arr2 = f(nums2, k - x);
            int[] arr = merge(arr1, arr2);
            if (compare(arr, ans, 0, 0)) {
                ans = arr;
            }
        }
        return ans;
    }

    private int[] f(int[] nums, int k) {
        int n = nums.length;
        int[] stk = new int[k];
        int top = -1;
        int remain = n - k;
        for (int x : nums) {
            while (top >= 0 && stk[top] < x && remain > 0) {
                --top;
                --remain;
            }
            if (top + 1 < k) {
                stk[++top] = x;
            } else {
                --remain;
            }
        }
        return stk;
    }

    private int[] merge(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int i = 0, j = 0;
        int[] ans = new int[m + n];
        for (int k = 0; k < m + n; ++k) {
            if (compare(nums1, nums2, i, j)) {
                ans[k] = nums1[i++];
            } else {
                ans[k] = nums2[j++];
            }
        }
        return ans;
    }

    private boolean compare(int[] nums1, int[] nums2, int i, int j) {
        if (i >= nums1.length) {
            return false;
        }
        if (j >= nums2.length) {
            return true;
        }
        if (nums1[i] > nums2[j]) {
            return true;
        }
        if (nums1[i] < nums2[j]) {
            return false;
        }
        return compare(nums1, nums2, i + 1, j + 1);
    }
}

// Accepted solution for LeetCode #321: Create Maximum Number
func maxNumber(nums1 []int, nums2 []int, k int) []int {
	m, n := len(nums1), len(nums2)
	l, r := max(0, k-n), min(k, m)
	f := func(nums []int, k int) []int {
		n := len(nums)
		stk := make([]int, k)
		top := -1
		remain := n - k
		for _, x := range nums {
			for top >= 0 && stk[top] < x && remain > 0 {
				top--
				remain--
			}
			if top+1 < k {
				top++
				stk[top] = x
			} else {
				remain--
			}
		}
		return stk
	}

	var compare func(nums1, nums2 []int, i, j int) bool
	compare = func(nums1, nums2 []int, i, j int) bool {
		if i >= len(nums1) {
			return false
		}
		if j >= len(nums2) {
			return true
		}
		if nums1[i] > nums2[j] {
			return true
		}
		if nums1[i] < nums2[j] {
			return false
		}
		return compare(nums1, nums2, i+1, j+1)
	}

	merge := func(nums1, nums2 []int) []int {
		m, n := len(nums1), len(nums2)
		ans := make([]int, m+n)
		i, j := 0, 0
		for k := range ans {
			if compare(nums1, nums2, i, j) {
				ans[k] = nums1[i]
				i++
			} else {
				ans[k] = nums2[j]
				j++
			}
		}
		return ans
	}

	ans := make([]int, k)
	for x := l; x <= r; x++ {
		arr1 := f(nums1, x)
		arr2 := f(nums2, k-x)
		arr := merge(arr1, arr2)
		if compare(arr, ans, 0, 0) {
			ans = arr
		}
	}
	return ans
}

# Accepted solution for LeetCode #321: Create Maximum Number
class Solution:
    def maxNumber(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:
        def f(nums: List[int], k: int) -> List[int]:
            n = len(nums)
            stk = [0] * k
            top = -1
            remain = n - k
            for x in nums:
                while top >= 0 and stk[top] < x and remain > 0:
                    top -= 1
                    remain -= 1
                if top + 1 < k:
                    top += 1
                    stk[top] = x
                else:
                    remain -= 1
            return stk

        def compare(nums1: List[int], nums2: List[int], i: int, j: int) -> bool:
            if i >= len(nums1):
                return False
            if j >= len(nums2):
                return True
            if nums1[i] > nums2[j]:
                return True
            if nums1[i] < nums2[j]:
                return False
            return compare(nums1, nums2, i + 1, j + 1)

        def merge(nums1: List[int], nums2: List[int]) -> List[int]:
            m, n = len(nums1), len(nums2)
            i = j = 0
            ans = [0] * (m + n)
            for k in range(m + n):
                if compare(nums1, nums2, i, j):
                    ans[k] = nums1[i]
                    i += 1
                else:
                    ans[k] = nums2[j]
                    j += 1
            return ans

        m, n = len(nums1), len(nums2)
        l, r = max(0, k - n), min(k, m)
        ans = [0] * k
        for x in range(l, r + 1):
            arr1 = f(nums1, x)
            arr2 = f(nums2, k - x)
            arr = merge(arr1, arr2)
            if ans < arr:
                ans = arr
        return ans

// Accepted solution for LeetCode #321: Create Maximum Number
struct Solution;

impl Solution {
    fn max_number(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> Vec<i32> {
        let n1 = nums1.len();
        let n2 = nums2.len();
        let k = k as usize;
        let mut max_merged: Option<Vec<i32>> = None;
        for size1 in 0..=k.min(n1) {
            let size2 = k - size1;
            if size2 > n2 {
                continue;
            }
            let max1 = Self::max_one(&nums1, size1);
            let max2 = Self::max_one(&nums2, size2);
            let max3 = Self::max_merge(max1, max2);
            if let Some(max) = max_merged {
                if max < max3 {
                    max_merged = Some(max3);
                } else {
                    max_merged = Some(max);
                }
            } else {
                max_merged = Some(max3);
            }
        }
        max_merged.unwrap()
    }

    fn max_one(nums: &[i32], k: usize) -> Vec<i32> {
        let mut stack = vec![];
        let n = nums.len();
        for i in 0..n {
            let right = n - i;
            while let Some(&top) = stack.last() {
                if top < nums[i] && stack.len() + right > k {
                    stack.pop();
                } else {
                    break;
                }
            }
            stack.push(nums[i]);
        }
        while stack.len() > k {
            stack.pop();
        }
        stack
    }

    fn max_merge(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut res = vec![];
        let mut i = 0;
        let mut j = 0;
        loop {
            if i < nums1.len() && j < nums2.len() {
                if Self::greater(&nums1, &nums2, i, j) {
                    res.push(nums1[i]);
                    i += 1;
                } else {
                    res.push(nums2[j]);
                    j += 1;
                }
                continue;
            }
            if i < nums1.len() {
                res.push(nums1[i]);
                i += 1;
                continue;
            }
            if j < nums2.len() {
                res.push(nums2[j]);
                j += 1;
                continue;
            }
            break;
        }
        res
    }

    fn greater(nums1: &[i32], nums2: &[i32], mut i: usize, mut j: usize) -> bool {
        while i < nums1.len() && j < nums2.len() && nums1[i] == nums2[j] {
            i += 1;
            j += 1;
        }
        j == nums2.len() || (i < nums1.len() && nums1[i] > nums2[j])
    }
}

#[test]
fn test() {
    let nums1 = vec![3, 4, 6, 5];
    let nums2 = vec![9, 1, 2, 5, 8, 3];
    let k = 5;
    let res = vec![9, 8, 6, 5, 3];
    assert_eq!(Solution::max_number(nums1, nums2, k), res);
    let nums1 = vec![6, 7];
    let nums2 = vec![6, 0, 4];
    let k = 5;
    let res = vec![6, 7, 6, 0, 4];
    assert_eq!(Solution::max_number(nums1, nums2, k), res);
    let nums1 = vec![3, 9];
    let nums2 = vec![8, 9];
    let k = 3;
    let res = vec![9, 8, 9];
    assert_eq!(Solution::max_number(nums1, nums2, k), res);
    let nums1 = vec![2, 5, 6, 4, 4, 0];
    let nums2 = vec![7, 3, 8, 0, 6, 5, 7, 6, 2];
    let k = 15;
    let res = vec![7, 3, 8, 2, 5, 6, 4, 4, 0, 6, 5, 7, 6, 2, 0];
    assert_eq!(Solution::max_number(nums1, nums2, k), res);
}

// Accepted solution for LeetCode #321: Create Maximum Number
function maxNumber(nums1: number[], nums2: number[], k: number): number[] {
    const m = nums1.length;
    const n = nums2.length;
    const l = Math.max(0, k - n);
    const r = Math.min(k, m);
    let ans: number[] = Array(k).fill(0);
    for (let x = l; x <= r; ++x) {
        const arr1 = f(nums1, x);
        const arr2 = f(nums2, k - x);
        const arr = merge(arr1, arr2);
        if (compare(arr, ans, 0, 0)) {
            ans = arr;
        }
    }
    return ans;
}

function f(nums: number[], k: number): number[] {
    const n = nums.length;
    const stk: number[] = Array(k).fill(0);
    let top = -1;
    let remain = n - k;
    for (const x of nums) {
        while (top >= 0 && stk[top] < x && remain > 0) {
            --top;
            --remain;
        }
        if (top + 1 < k) {
            stk[++top] = x;
        } else {
            --remain;
        }
    }
    return stk;
}

function compare(nums1: number[], nums2: number[], i: number, j: number): boolean {
    if (i >= nums1.length) {
        return false;
    }
    if (j >= nums2.length) {
        return true;
    }
    if (nums1[i] > nums2[j]) {
        return true;
    }
    if (nums1[i] < nums2[j]) {
        return false;
    }
    return compare(nums1, nums2, i + 1, j + 1);
}

function merge(nums1: number[], nums2: number[]): number[] {
    const m = nums1.length;
    const n = nums2.length;
    const ans: number[] = Array(m + n).fill(0);
    let i = 0;
    let j = 0;
    for (let k = 0; k < m + n; ++k) {
        if (compare(nums1, nums2, i, j)) {
            ans[k] = nums1[i++];
        } else {
            ans[k] = nums2[j++];
        }
    }
    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 of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.