LeetCode #3523 — MEDIUM

Make Array Non-decreasing

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

The Problem

Problem Statement

You are given an integer array nums. In one operation, you can select a subarray and replace it with a single element equal to its maximum value.

Return the maximum possible size of the array after performing zero or more operations such that the resulting array is non-decreasing.

Example 1:

Input: nums = [4,2,5,3,5]

Output: 3

Explanation:

One way to achieve the maximum size is:

Replace subarray nums[1..2] = [2, 5] with 5 → [4, 5, 3, 5].
Replace subarray nums[2..3] = [3, 5] with 5 → [4, 5, 5].

The final array [4, 5, 5] is non-decreasing with size 3.

Example 2:

Input: nums = [1,2,3]

Output: 3

Explanation:

No operation is needed as the array [1,2,3] is already non-decreasing.

Constraints:

1 <= nums.length <= 2 * 10⁵
1 <= nums[i] <= 2 * 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. In one operation, you can select a subarray and replace it with a single element equal to its maximum value. Return the maximum possible size of the array after performing zero or more operations such that the resulting array is non-decreasing.

Baseline thinking

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

Pattern signal: Array · Stack · Greedy

Example 1

[4,2,5,3,5]

Example 2

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

Iterate backwards.
Can you remove the largest element in the array? Is that ever helpful?

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 #3523: Make Array Non-decreasing
class Solution {
    public int maximumPossibleSize(int[] nums) {
        int ans = 0, mx = 0;
        for (int x : nums) {
            if (mx <= x) {
                ++ans;
                mx = x;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3523: Make Array Non-decreasing
func maximumPossibleSize(nums []int) int {
	ans, mx := 0, 0
	for _, x := range nums {
		if mx <= x {
			ans++
			mx = x
		}
	}
	return ans
}

# Accepted solution for LeetCode #3523: Make Array Non-decreasing
class Solution:
    def maximumPossibleSize(self, nums: List[int]) -> int:
        ans = mx = 0
        for x in nums:
            if mx <= x:
                ans += 1
                mx = x
        return ans

// Accepted solution for LeetCode #3523: Make Array Non-decreasing
fn maximum_possible_size(nums: Vec<i32>) -> i32 {
    let mut ret = 0;
    let mut prev = 0;

    for num in nums {
        if num >= prev {
            prev = num;
            ret += 1;
        }
    }

    ret
}

fn main() {
    let nums = vec![4, 2, 5, 3, 5];
    let ret = maximum_possible_size(nums);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let nums = vec![4, 2, 5, 3, 5];
        let ret = maximum_possible_size(nums);
        assert_eq!(ret, 3);
    }
    {
        let nums = vec![1, 2, 3];
        let ret = maximum_possible_size(nums);
        assert_eq!(ret, 3);
    }
}

// Accepted solution for LeetCode #3523: Make Array Non-decreasing
function maximumPossibleSize(nums: number[]): number {
    let [ans, mx] = [0, 0];
    for (const x of nums) {
        if (mx <= x) {
            ++ans;
            mx = x;
        }
    }
    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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

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.