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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You are given a 0-indexed permutation of n integers nums.
A permutation is called semi-ordered if the first number equals 1 and the last number equals n. You can perform the below operation as many times as you want until you make nums a semi-ordered permutation:
nums, then swap them.Return the minimum number of operations to make nums a semi-ordered permutation.
A permutation is a sequence of integers from 1 to n of length n containing each number exactly once.
Example 1:
Input: nums = [2,1,4,3] Output: 2 Explanation: We can make the permutation semi-ordered using these sequence of operations: 1 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3]. 2 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4]. It can be proved that there is no sequence of less than two operations that make nums a semi-ordered permutation.
Example 2:
Input: nums = [2,4,1,3] Output: 3 Explanation: We can make the permutation semi-ordered using these sequence of operations: 1 - swap i = 1 and j = 2. The permutation becomes [2,1,4,3]. 2 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3]. 3 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4]. It can be proved that there is no sequence of less than three operations that make nums a semi-ordered permutation.
Example 3:
Input: nums = [1,3,4,2,5] Output: 0 Explanation: The permutation is already a semi-ordered permutation.
Constraints:
2 <= nums.length == n <= 501 <= nums[i] <= 50nums is a permutation.Problem summary: You are given a 0-indexed permutation of n integers nums. A permutation is called semi-ordered if the first number equals 1 and the last number equals n. You can perform the below operation as many times as you want until you make nums a semi-ordered permutation: Pick two adjacent elements in nums, then swap them. Return the minimum number of operations to make nums a semi-ordered permutation. A permutation is a sequence of integers from 1 to n of length n containing each number exactly once.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[2,1,4,3]
[2,4,1,3]
[1,3,4,2,5]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2717: Semi-Ordered Permutation
class Solution {
public int semiOrderedPermutation(int[] nums) {
int n = nums.length;
int i = 0, j = 0;
for (int k = 0; k < n; ++k) {
if (nums[k] == 1) {
i = k;
}
if (nums[k] == n) {
j = k;
}
}
int k = i < j ? 1 : 2;
return i + n - j - k;
}
}
// Accepted solution for LeetCode #2717: Semi-Ordered Permutation
func semiOrderedPermutation(nums []int) int {
n := len(nums)
var i, j int
for k, x := range nums {
if x == 1 {
i = k
}
if x == n {
j = k
}
}
k := 1
if i > j {
k = 2
}
return i + n - j - k
}
# Accepted solution for LeetCode #2717: Semi-Ordered Permutation
class Solution:
def semiOrderedPermutation(self, nums: List[int]) -> int:
n = len(nums)
i = nums.index(1)
j = nums.index(n)
k = 1 if i < j else 2
return i + n - j - k
// Accepted solution for LeetCode #2717: Semi-Ordered Permutation
impl Solution {
pub fn semi_ordered_permutation(nums: Vec<i32>) -> i32 {
let n = nums.len();
let (mut i, mut j) = (0, 0);
for k in 0..n {
if nums[k] == 1 {
i = k;
}
if nums[k] == (n as i32) {
j = k;
}
}
let k = if i < j { 1 } else { 2 };
(i + n - j - k) as i32
}
}
// Accepted solution for LeetCode #2717: Semi-Ordered Permutation
function semiOrderedPermutation(nums: number[]): number {
const n = nums.length;
const i = nums.indexOf(1);
const j = nums.indexOf(n);
const k = i < j ? 1 : 2;
return i + n - j - k;
}
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.