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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given an integer array nums.
You need to remove exactly one prefix (possibly empty) from nums.
Return an integer denoting the minimum length of the removed prefix such that the remaining array is strictly increasing.
Example 1:
Input: nums = [1,-1,2,3,3,4,5]
Output: 4
Explanation:
Removing the prefix = [1, -1, 2, 3] leaves the remaining array [3, 4, 5] which is strictly increasing.
Example 2:
Input: nums = [4,3,-2,-5]
Output: 3
Explanation:
Removing the prefix = [4, 3, -2] leaves the remaining array [-5] which is strictly increasing.
Example 3:
Input: nums = [1,2,3,4]
Output: 0
Explanation:
The array nums = [1, 2, 3, 4] is already strictly increasing so removing an empty prefix is sufficient.
Constraints:
1 <= nums.length <= 105-109 <= nums[i] <= 109Problem summary: You are given an integer array nums. You need to remove exactly one prefix (possibly empty) from nums. Return an integer denoting the minimum length of the removed prefix such that the remaining array is strictly increasing.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[1,-1,2,3,3,4,5]
[4,3,-2,-5]
[1,2,3,4]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3818: Minimum Prefix Removal to Make Array Strictly Increasing
class Solution {
public int minimumPrefixLength(int[] nums) {
for (int i = nums.length - 1; i > 0; --i) {
if (nums[i - 1] >= nums[i]) {
return i;
}
}
return 0;
}
}
// Accepted solution for LeetCode #3818: Minimum Prefix Removal to Make Array Strictly Increasing
func minimumPrefixLength(nums []int) int {
for i := len(nums) - 1; i > 0; i-- {
if nums[i-1] >= nums[i] {
return i
}
}
return 0
}
# Accepted solution for LeetCode #3818: Minimum Prefix Removal to Make Array Strictly Increasing
class Solution:
def minimumPrefixLength(self, nums: List[int]) -> int:
for i in range(len(nums) - 1, 0, -1):
if nums[i - 1] >= nums[i]:
return i
return 0
// Accepted solution for LeetCode #3818: Minimum Prefix Removal to Make Array Strictly Increasing
fn minimum_prefix_length(nums: Vec<i32>) -> i32 {
let len = nums.len();
for i in (1..len).rev() {
if nums[i - 1] >= nums[i] {
return i as i32;
}
}
0
}
fn main() {
let ret = minimum_prefix_length(vec![1, -1, 2, 3, 3, 4, 5]);
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(minimum_prefix_length(vec![1, -1, 2, 3, 3, 4, 5]), 4);
assert_eq!(minimum_prefix_length(vec![4, 3, -2, -5]), 3);
assert_eq!(minimum_prefix_length(vec![1, 2, 3, 4]), 0);
}
// Accepted solution for LeetCode #3818: Minimum Prefix Removal to Make Array Strictly Increasing
function minimumPrefixLength(nums: number[]): number {
for (let i = nums.length - 1; i; --i) {
if (nums[i - 1] >= nums[i]) {
return i;
}
}
return 0;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.