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You are given an integer array nums and an integer k.
Initially, you have k units of resources.
You must process the elements of nums from left to right. To process the ith element, you need nums[i] resources.
If your available resources are less than nums[i], you may perform an operation that increases your available resources by k. The value of k is fixed and does not change throughout the process. The first such operation incurs a cost of 1, the second incurs a cost of 2, and so on.
After processing the ith element, your available resources decrease by nums[i].
Return an integer denoting the minimum total cost required to process all elements. Since the answer may be very large, return it modulo 109 + 7.
Example 1:
Input: nums = [1,2,3,4], k = 4
Output: 3
Explanation:
nums[0], we have 4 - 1 = 3 units of resources left.nums[1], we have 3 - 2 = 1 unit of resources left.nums[2] = 3 and only 1 unit of resources is available, we perform the first operation costing 1. After processing nums[2], we have 1 + 4 - 3 = 2 units of resources left.nums[3] = 4 and only 2 units of resources are available, we perform the second operation costing 2, to have 2 + 4 = 6 units of resources, which is enough to process nums[3].1 + 2 = 3.Example 2:
Input: nums = [1,1,7,14], k = 4
Output: 15
Explanation:
nums[0], we have 4 - 1 = 3 units of resources left.nums[1], we have 3 - 1 = 2 units of resources left.nums[2] = 7 and only 2 units of resources are available, we perform two operations costing 1 + 2 = 3. After processing nums[2], we have 2 + 4 + 4 - 7 = 3 units of resources left.nums[3] = 14 and only 3 units of resources are available, we perform three operations costing 3 + 4 + 5 = 12, to have 3 + 4 + 4 + 4 = 15 units of resources, which is enough to process nums[3].3 + 12 = 15.Example 3:
Input: nums = [1,2,3,4], k = 10
Output: 0
Explanation:
To process all elements, we can use the initial 10 units of resources without performing any operations. Thus, the total cost required is 0.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 109Problem Overview: You need to process every element while minimizing the total accumulated cost under a set of transition or operation rules. The challenge is choosing the processing order or state transition that avoids expensive repeated operations and keeps the running cost minimal.
Approach 1: Brute Force Recursion (O(2^n) time, O(n) space)
The direct approach tries every possible decision at each element. You recursively decide whether to process the current value immediately, combine it with another operation, or defer the cost depending on the problem constraints. This works for very small inputs because it explores the full decision tree, but overlapping subproblems quickly make it impractical. Use this version only to validate transitions before optimizing.
Approach 2: Top-Down Dynamic Programming (O(n^2) time, O(n) space)
Memoization removes repeated work by caching the minimum cost for each state. The key insight is that once you know the optimal cost from index i onward, you never need to recompute it again. Most solutions define a DP state around the current position and the previous processing choice, then transition using the cheapest valid operation. This is the standard optimization for problems involving sequential decisions and cumulative costs. Related concepts appear frequently in dynamic programming and arrays problems.
Approach 3: Greedy + Priority Structure Optimization (O(n log n) time, O(n) space)
The optimal implementation usually combines a greedy observation with a supporting data structure such as a heap or ordered container. Instead of evaluating every possible transition, you keep track of the currently cheapest processing option and update it as you iterate through the array. Heap insertion and removal stay logarithmic, which reduces the overall runtime significantly for large inputs. This approach performs well when local minimum choices lead to a globally optimal answer, especially in scheduling-style problems linked to greedy algorithms.
Approach 4: Bottom-Up DP with Prefix Optimization (O(n) or O(n log n) time, O(n) space)
Some implementations avoid recursion entirely by building the answer iteratively from left to right. Prefix minimums, monotonic transitions, or rolling-state compression reduce the transition cost from quadratic to linear or near-linear time. You update the best achievable cost for each position using previously computed states instead of restarting calculations repeatedly. This version is preferred in production code because it avoids recursion overhead and scales better under strict constraints.
Recommended for interviews: Start with the brute force recursion to explain the decision process clearly. Then optimize it into memoized DP to show recognition of overlapping subproblems. Interviewers usually expect the heap-optimized or bottom-up DP solution because it demonstrates control over both algorithmic complexity and implementation detail.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Recursion | O(2^n) | O(n) | Useful for understanding transitions or validating small test cases |
| Top-Down Dynamic Programming | O(n^2) | O(n) | General-purpose optimization for overlapping subproblems |
| Greedy + Heap Optimization | O(n log n) | O(n) | Best when the cheapest local choice can be tracked dynamically |
| Bottom-Up DP with Prefix Optimization | O(n) to O(n log n) | O(n) | Preferred for large constraints and iterative implementations |