LeetCode #1909 — EASY

Remove One Element to Make the Array Strictly Increasing

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given a 0-indexed integer array nums, return true if it can be made strictly increasing after removing exactly one element, or false otherwise. If the array is already strictly increasing, return true.

The array nums is strictly increasing if nums[i - 1] < nums[i] for each index (1 <= i < nums.length).

Example 1:

Input: nums = [1,2,10,5,7]
Output: true
Explanation: By removing 10 at index 2 from nums, it becomes [1,2,5,7].
[1,2,5,7] is strictly increasing, so return true.

Example 2:

Input: nums = [2,3,1,2]
Output: false
Explanation:
[3,1,2] is the result of removing the element at index 0.
[2,1,2] is the result of removing the element at index 1.
[2,3,2] is the result of removing the element at index 2.
[2,3,1] is the result of removing the element at index 3.
No resulting array is strictly increasing, so return false.

Example 3:

Input: nums = [1,1,1]
Output: false
Explanation: The result of removing any element is [1,1].
[1,1] is not strictly increasing, so return false.

Constraints:

2 <= nums.length <= 1000
1 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: Given a 0-indexed integer array nums, return true if it can be made strictly increasing after removing exactly one element, or false otherwise. If the array is already strictly increasing, return true. The array nums is strictly increasing if nums[i - 1] < nums[i] for each index (1 <= i < nums.length).

Baseline thinking

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

Pattern signal: Array

Example 1

[1,2,10,5,7]

Example 2

[2,3,1,2]

Example 3

[1,1,1]

Core Insight

What unlocks the optimal approach

For each index i in nums remove this index.
If the array becomes sorted return true, otherwise revert to the original array and try different index.

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 #1909: Remove One Element to Make the Array Strictly Increasing
class Solution {
    public boolean canBeIncreasing(int[] nums) {
        int i = 0;
        while (i + 1 < nums.length && nums[i] < nums[i + 1]) {
            ++i;
        }
        return check(nums, i) || check(nums, i + 1);
    }

    private boolean check(int[] nums, int k) {
        int pre = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (i == k) {
                continue;
            }
            if (pre >= nums[i]) {
                return false;
            }
            pre = nums[i];
        }
        return true;
    }
}

// Accepted solution for LeetCode #1909: Remove One Element to Make the Array Strictly Increasing
func canBeIncreasing(nums []int) bool {
	check := func(k int) bool {
		pre := 0
		for i, x := range nums {
			if i == k {
				continue
			}
			if pre >= x {
				return false
			}
			pre = x
		}
		return true
	}
	i := 0
	for i+1 < len(nums) && nums[i] < nums[i+1] {
		i++
	}
	return check(i) || check(i+1)
}

# Accepted solution for LeetCode #1909: Remove One Element to Make the Array Strictly Increasing
class Solution:
    def canBeIncreasing(self, nums: List[int]) -> bool:
        def check(k: int) -> bool:
            pre = -inf
            for i, x in enumerate(nums):
                if i == k:
                    continue
                if pre >= x:
                    return False
                pre = x
            return True

        i = 0
        while i + 1 < len(nums) and nums[i] < nums[i + 1]:
            i += 1
        return check(i) or check(i + 1)

// Accepted solution for LeetCode #1909: Remove One Element to Make the Array Strictly Increasing
impl Solution {
    pub fn can_be_increasing(nums: Vec<i32>) -> bool {
        let check = |k: usize| -> bool {
            let mut pre = 0;
            for (i, &x) in nums.iter().enumerate() {
                if i == k {
                    continue;
                }
                if pre >= x {
                    return false;
                }
                pre = x;
            }
            true
        };

        let mut i = 0;
        while i + 1 < nums.len() && nums[i] < nums[i + 1] {
            i += 1;
        }
        check(i) || check(i + 1)
    }
}

// Accepted solution for LeetCode #1909: Remove One Element to Make the Array Strictly Increasing
function canBeIncreasing(nums: number[]): boolean {
    const n = nums.length;
    const check = (k: number): boolean => {
        let pre = 0;
        for (let i = 0; i < n; ++i) {
            if (i === k) {
                continue;
            }
            if (pre >= nums[i]) {
                return false;
            }
            pre = nums[i];
        }
        return true;
    };
    let i = 0;
    while (i + 1 < n && nums[i] < nums[i + 1]) {
        ++i;
    }
    return check(i) || check(i + 1);
}

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 or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.