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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given an integer array rolls of length n and an integer k. You roll a k sided dice numbered from 1 to k, n times, where the result of the ith roll is rolls[i].
Return the length of the shortest sequence of rolls so that there's no such subsequence in rolls.
A sequence of rolls of length len is the result of rolling a k sided dice len times.
Example 1:
Input: rolls = [4,2,1,2,3,3,2,4,1], k = 4 Output: 3 Explanation: Every sequence of rolls of length 1, [1], [2], [3], [4], can be taken from rolls. Every sequence of rolls of length 2, [1, 1], [1, 2], ..., [4, 4], can be taken from rolls. The sequence [1, 4, 2] cannot be taken from rolls, so we return 3. Note that there are other sequences that cannot be taken from rolls.
Example 2:
Input: rolls = [1,1,2,2], k = 2 Output: 2 Explanation: Every sequence of rolls of length 1, [1], [2], can be taken from rolls. The sequence [2, 1] cannot be taken from rolls, so we return 2. Note that there are other sequences that cannot be taken from rolls but [2, 1] is the shortest.
Example 3:
Input: rolls = [1,1,3,2,2,2,3,3], k = 4 Output: 1 Explanation: The sequence [4] cannot be taken from rolls, so we return 1. Note that there are other sequences that cannot be taken from rolls but [4] is the shortest.
Constraints:
n == rolls.length1 <= n <= 1051 <= rolls[i] <= k <= 105Problem summary: You are given an integer array rolls of length n and an integer k. You roll a k sided dice numbered from 1 to k, n times, where the result of the ith roll is rolls[i]. Return the length of the shortest sequence of rolls so that there's no such subsequence in rolls. A sequence of rolls of length len is the result of rolling a k sided dice len times.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Greedy
[4,2,1,2,3,3,2,4,1] 4
[1,1,2,2] 2
[1,1,3,2,2,2,3,3] 4
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
class Solution {
public int shortestSequence(int[] rolls, int k) {
Set<Integer> s = new HashSet<>();
int ans = 1;
for (int v : rolls) {
s.add(v);
if (s.size() == k) {
s.clear();
++ans;
}
}
return ans;
}
}
// Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
func shortestSequence(rolls []int, k int) int {
s := map[int]bool{}
ans := 1
for _, v := range rolls {
s[v] = true
if len(s) == k {
ans++
s = map[int]bool{}
}
}
return ans
}
# Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
class Solution:
def shortestSequence(self, rolls: List[int], k: int) -> int:
ans = 1
s = set()
for v in rolls:
s.add(v)
if len(s) == k:
ans += 1
s.clear()
return ans
// Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
/**
* [2350] Shortest Impossible Sequence of Rolls
*
* You are given an integer array rolls of length n and an integer k. You roll a k sided dice numbered from 1 to k, n times, where the result of the i^th roll is rolls[i].
* Return the length of the shortest sequence of rolls so that there's no such <span data-keyword="subsequence-array">subsequence</span> in rolls.
* A sequence of rolls of length len is the result of rolling a k sided dice len times.
*
* Example 1:
*
* Input: rolls = [4,2,1,2,3,3,2,4,1], k = 4
* Output: 3
* Explanation: Every sequence of rolls of length 1, [1], [2], [3], [4], can be taken from rolls.
* Every sequence of rolls of length 2, [1, 1], [1, 2], ..., [4, 4], can be taken from rolls.
* The sequence [1, 4, 2] cannot be taken from rolls, so we return 3.
* Note that there are other sequences that cannot be taken from rolls.
* Example 2:
*
* Input: rolls = [1,1,2,2], k = 2
* Output: 2
* Explanation: Every sequence of rolls of length 1, [1], [2], can be taken from rolls.
* The sequence [2, 1] cannot be taken from rolls, so we return 2.
* Note that there are other sequences that cannot be taken from rolls but [2, 1] is the shortest.
*
* Example 3:
*
* Input: rolls = [1,1,3,2,2,2,3,3], k = 4
* Output: 1
* Explanation: The sequence [4] cannot be taken from rolls, so we return 1.
* Note that there are other sequences that cannot be taken from rolls but [4] is the shortest.
*
*
* Constraints:
*
* n == rolls.length
* 1 <= n <= 10^5
* 1 <= rolls[i] <= k <= 10^5
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/shortest-impossible-sequence-of-rolls/
// discuss: https://leetcode.com/problems/shortest-impossible-sequence-of-rolls/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn shortest_sequence(rolls: Vec<i32>, k: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_2350_example_1() {
let rolls = vec![4, 2, 1, 2, 3, 3, 2, 4, 1];
let k = 4;
let result = 3;
assert_eq!(Solution::shortest_sequence(rolls, k), result);
}
#[test]
#[ignore]
fn test_2350_example_2() {
let rolls = vec![1, 1, 2, 2];
let k = 2;
let result = 2;
assert_eq!(Solution::shortest_sequence(rolls, k), result);
}
#[test]
#[ignore]
fn test_2350_example_3() {
let rolls = vec![1, 1, 3, 2, 2, 2, 3, 3];
let k = 4;
let result = 1;
assert_eq!(Solution::shortest_sequence(rolls, k), result);
}
}
// Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2350: Shortest Impossible Sequence of Rolls
// class Solution {
// public int shortestSequence(int[] rolls, int k) {
// Set<Integer> s = new HashSet<>();
// int ans = 1;
// for (int v : rolls) {
// s.add(v);
// if (s.size() == k) {
// s.clear();
// ++ans;
// }
// }
// return ans;
// }
// }
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
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.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
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.