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 a 0-indexed integer array nums having length n, and an integer k.
You can perform the following increment operation any number of times (including zero):
i in the range [0, n - 1], and increase nums[i] by 1.An array is considered beautiful if, for any subarray with a size of 3 or more, its maximum element is greater than or equal to k.
Return an integer denoting the minimum number of increment operations needed to make nums beautiful.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [2,3,0,0,2], k = 4 Output: 3 Explanation: We can perform the following increment operations to make nums beautiful: Choose index i = 1 and increase nums[1] by 1 -> [2,4,0,0,2]. Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,3]. Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,4]. The subarrays with a size of 3 or more are: [2,4,0], [4,0,0], [0,0,4], [2,4,0,0], [4,0,0,4], [2,4,0,0,4]. In all the subarrays, the maximum element is equal to k = 4, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 3 increment operations. Hence, the answer is 3.
Example 2:
Input: nums = [0,1,3,3], k = 5 Output: 2 Explanation: We can perform the following increment operations to make nums beautiful: Choose index i = 2 and increase nums[2] by 1 -> [0,1,4,3]. Choose index i = 2 and increase nums[2] by 1 -> [0,1,5,3]. The subarrays with a size of 3 or more are: [0,1,5], [1,5,3], [0,1,5,3]. In all the subarrays, the maximum element is equal to k = 5, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 2 increment operations. Hence, the answer is 2.
Example 3:
Input: nums = [1,1,2], k = 1 Output: 0 Explanation: The only subarray with a size of 3 or more in this example is [1,1,2]. The maximum element, 2, is already greater than k = 1, so we don't need any increment operation. Hence, the answer is 0.
Constraints:
3 <= n == nums.length <= 1050 <= nums[i] <= 1090 <= k <= 109Problem summary: You are given a 0-indexed integer array nums having length n, and an integer k. You can perform the following increment operation any number of times (including zero): Choose an index i in the range [0, n - 1], and increase nums[i] by 1. An array is considered beautiful if, for any subarray with a size of 3 or more, its maximum element is greater than or equal to k. Return an integer denoting the minimum number of increment operations needed to make nums beautiful. A subarray is a contiguous non-empty sequence of elements within an array.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming
[2,3,0,0,2] 4
[0,1,3,3] 5
[1,1,2] 1
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
class Solution {
public long minIncrementOperations(int[] nums, int k) {
long f = 0, g = 0, h = 0;
for (int x : nums) {
long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
f = g;
g = h;
h = hh;
}
return Math.min(Math.min(f, g), h);
}
}
// Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
func minIncrementOperations(nums []int, k int) int64 {
var f, g, h int
for _, x := range nums {
f, g, h = g, h, min(f, g, h)+max(k-x, 0)
}
return int64(min(f, g, h))
}
# Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
class Solution:
def minIncrementOperations(self, nums: List[int], k: int) -> int:
f = g = h = 0
for x in nums:
f, g, h = g, h, min(f, g, h) + max(k - x, 0)
return min(f, g, h)
// Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
// class Solution {
// public long minIncrementOperations(int[] nums, int k) {
// long f = 0, g = 0, h = 0;
// for (int x : nums) {
// long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
// f = g;
// g = h;
// h = hh;
// }
// return Math.min(Math.min(f, g), h);
// }
// }
// Accepted solution for LeetCode #2919: Minimum Increment Operations to Make Array Beautiful
function minIncrementOperations(nums: number[], k: number): number {
let [f, g, h] = [0, 0, 0];
for (const x of nums) {
[f, g, h] = [g, h, Math.min(f, g, h) + Math.max(k - x, 0)];
}
return Math.min(f, g, h);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
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: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.