Minimum Operations to Achieve At Least K Peaks - Video Solutions
Super Hard DP DSA Question - LeetCode Weekly Contest 496 - Q4 - (3892) Explained By Kumar K sir
Minimum Operations to Achieve At Least K Peaks - Video Solution
Watch 5 video solutions for Minimum Operations to Achieve At Least K Peaks, a hard level problem involving Array, Dynamic Programming. This walkthrough by Kumar K [Amazon] has 615 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
You are given a ​​​​​​​circular integer array​​​​​​​ nums of length n.
An index i is a peak if its value is strictly greater than its neighbors:
- The previous neighbor of
iisnums[i - 1]ifi > 0, otherwisenums[n - 1]. - The next neighbor of
iisnums[i + 1]ifi < n - 1, otherwisenums[0].
You are allowed to perform the following operation any number of times:
- Choose any index
iand increasenums[i]by 1.
Return an integer denoting the minimum number of operations required to make the array contain at least k peaks. If it is impossible, return -1.
Example 1:
Input: nums = [2,1,2], k = 1
Output: 1
Explanation:
- To achieve at least
k = 1peak, we can increasenums[2] = 2to 3. - After this operation,
nums[2] = 3is strictly greater than its neighborsnums[0] = 2andnums[1] = 1. - Therefore, the minimum number of operations required is 1.
Example 2:
Input: nums = [4,5,3,6], k = 2
Output: 0
Explanation:
- The array already contains at least
k = 2peaks with zero operations. - Index 1:
nums[1] = 5is strictly greater than its neighborsnums[0] = 4andnums[2] = 3. - Index 3:
nums[3] = 6is strictly greater than its neighborsnums[2] = 3andnums[0] = 4. - Therefore, the minimum number of operations required is 0.
Example 3:
Input: nums = [3,7,3], k = 2
Output: -1
Explanation:
It is impossible to have at least k = 2 peaks in this array. Therefore, the answer is -1.
Constraints:
2 <= n == nums.length <= 5000-105 <= nums[i] <= 1050 <= k <= n​​​​​​​
Approach Overview
Problem Overview: You receive an array and must perform operations so that the array contains at least k peaks. An index i is a peak when nums[i] > nums[i-1] and nums[i] > nums[i+1]. Each operation modifies array values, and the goal is to achieve at least k valid peaks with the minimum total cost.
Approach 1: Brute Force Peak Selection (Exponential)
Enumerate all subsets of indices that could serve as peaks. For every candidate set of size k, verify the non‑adjacency constraint and compute the cost required to raise each chosen index above its neighbors. The cost for index i is max(0, max(nums[i-1], nums[i+1]) - nums[i] + 1). Track the minimum cost among all valid selections. This approach demonstrates the peak condition clearly but runs in O(2^n) time with O(1) extra space, which becomes infeasible once n grows.
Approach 2: Dynamic Programming with Peak Placement (O(n*k))
Model the problem as choosing positions for peaks while avoiding adjacent placements. Define dp[i][j] as the minimum cost to process the first i elements and create j peaks where the i-th index is considered as a candidate. At each index you either skip it or turn it into a peak. When selecting i as a peak, compute the adjustment cost needed to make nums[i] greater than both neighbors. Because peaks cannot be adjacent, the transition comes from i-2. This structured choice turns the combinatorial search into a predictable table fill using dynamic programming. Time complexity is O(n*k) and space complexity is O(n*k).
Approach 3: Space-Optimized DP (O(n*k) time, O(k) space)
The DP only depends on earlier rows (i-1 and i-2). Compress the table into rolling arrays that track the minimum cost for each peak count. Precompute the cost of turning each index into a peak, then update states while iterating through the array. This reduces memory usage while preserving the same recurrence. The logic still relies on careful index transitions typical in array and dynamic programming problems. Time complexity remains O(n*k) with O(k) space.
Recommended for interviews: The dynamic programming formulation is what interviewers usually expect. Starting with the brute force explanation shows you understand the peak constraints and cost calculation. Transitioning to the O(n*k) DP demonstrates the key insight: peak positions cannot be adjacent, so each decision only depends on earlier valid states.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Peak Selection | O(2^n) | O(1) | Conceptual understanding of peak constraints or very small arrays |
| Dynamic Programming with Peak Placement | O(n*k) | O(n*k) | General solution for large arrays where peak positions must be optimized |
| Space Optimized DP | O(n*k) | O(k) | Memory constrained environments or production implementations |