Minimum Total Cost to Process All Elements solution and explanation
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The problem
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:
- After processing
nums[0], we have4 - 1 = 3units of resources left. - After processing
nums[1], we have3 - 2 = 1unit of resources left. - Since
nums[2] = 3and only 1 unit of resources is available, we perform the first operation costing 1. After processingnums[2], we have1 + 4 - 3 = 2units of resources left. - Since
nums[3] = 4and only 2 units of resources are available, we perform the second operation costing 2, to have2 + 4 = 6units of resources, which is enough to processnums[3]. - Thus, the total cost is
1 + 2 = 3.
Example 2:
Input: nums = [1,1,7,14], k = 4
Output: 15
Explanation:
- After processing
nums[0], we have4 - 1 = 3units of resources left. - After processing
nums[1], we have3 - 1 = 2units of resources left. - Since
nums[2] = 7and only 2 units of resources are available, we perform two operations costing1 + 2 = 3. After processingnums[2], we have2 + 4 + 4 - 7 = 3units of resources left. - Since
nums[3] = 14and only 3 units of resources are available, we perform three operations costing3 + 4 + 5 = 12, to have3 + 4 + 4 + 4 = 15units of resources, which is enough to processnums[3]. - Thus, the total cost is
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 <= 109
Approach overview
Problem 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.
Solution
The i-th operation costs i, so if we perform cnt operations in total, the total cost is 1 + 2 + cdots + cnt = \dfrac{cnt(cnt+1)}{2}. Minimizing the total cost is equivalent to minimizing the number of operations.
Simulate the process from left to right. Maintain the current available resources cur (initially k) and the number of operations performed cnt. When processing an element x:
- If
cur \ge x, simply subtractx; - Otherwise, perform
m = \left\lceil\dfrac{x - cur}{k}\right\rceilmore operations to increase resources bym times k, then subtractx.
After the traversal, compute the triangular number of cnt and take the result modulo 10^9 + 7.
The time complexity is O(n), and the space complexity is O(1), where n is the length of the array nums.
class Solution:
def minimumCost(self, nums: list[int], k: int) -> int:
cnt = 0
cur = k
mod = 10**9 + 7
for x in nums:
diff = x - cur
if diff > 0:
m = (diff + k - 1) // k
cur += m * k
cnt += m
cur -= x
cnt %= mod
return (1 + cnt) * cnt // 2 % modclass Solution {
public int minimumCost(int[] nums, int k) {
final int MOD = 1_000_000_007;
long cnt = 0;
long cur = k;
for (int x : nums) {
long diff = (long) x - cur;
if (diff > 0) {
long m = (diff + k - 1L) / k;
cur += m * (long) k;
cnt += m;
}
cur -= x;
}
cnt %= MOD;
return (int) ((cnt + 1) * cnt / 2 % MOD);
}
}class Solution {
public:
int minimumCost(vector<int>& nums, int k) {
const int MOD = 1'000'000'007;
long long cnt = 0;
long long cur = k;
for (int x : nums) {
long long diff = (long long) x - cur;
if (diff > 0) {
long long m = (diff + k - 1LL) / k;
cur += m * k;
cnt += m;
}
cur -= x;
}
cnt %= MOD;
return (cnt + 1) * cnt / 2 % MOD;
}
};function minimumCost(nums: number[], k: number): number {
const MOD = 1000000007n;
let cnt = 0n;
let cur = BigInt(k);
const K = BigInt(k);
for (const x of nums) {
const diff = BigInt(x) - cur;
if (diff > 0n) {
const m = (diff + K - 1n) / K;
cur += m * K;
cnt += m;
}
cur -= BigInt(x);
}
cnt %= MOD;
return Number((((cnt + 1n) * cnt) / 2n) % MOD);
}func minimumCost(nums []int, k int) int {
const mod int64 = 1_000_000_007
var cnt int64
cur := int64(k)
for _, x := range nums {
diff := int64(x) - cur
if diff > 0 {
m := (diff + int64(k) - 1) / int64(k)
cur += m * int64(k)
cnt += m
}
cur -= int64(x)
}
cnt %= mod
return int((cnt + 1) * cnt / 2 % mod)
}Complexity analysis
| 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 |
Frequently asked questions
Is Minimum Total Cost to Process All Elements easy or hard?
Minimum Total Cost to Process All Elements is generally considered a medium-level problem because the main difficulty comes from recognizing the correct optimization strategy. The implementation is manageable once the DP state and transition logic are defined clearly.
Minimum Total Cost to Process All Elements Python/Java solution
Python solutions typically use memoization with functools.cache or heapq for priority queue optimization. Java implementations often rely on PriorityQueue and iterative DP arrays for better runtime consistency and memory control.
How to solve Minimum Total Cost to Process All Elements in O(n)?
An O(n) solution is possible when the transition between states can be simplified using prefix minimums, rolling DP states, or monotonic properties. Instead of recomputing costs for every previous index, you maintain the best candidate while iterating once through the array.
What is the best approach for Minimum Total Cost to Process All Elements?
The strongest approach combines dynamic programming with greedy optimization or a heap-based transition structure. It reduces repeated computations and keeps the minimum processing cost updated efficiently. Most accepted solutions run in O(n log n) or better depending on the transition rules.
Is Minimum Total Cost to Process All Elements asked at Google/Amazon/Meta?
Problems involving minimum processing cost, dynamic transitions, and greedy scheduling patterns appear frequently in Google, Amazon, and Meta interview rounds. Variants usually test state optimization, heap usage, and DP transition design under large constraints.
What data structure is used in Minimum Total Cost to Process All Elements?
Most optimized solutions use arrays for DP storage and heaps or priority queues for fast minimum-cost retrieval. Some implementations also rely on prefix arrays, ordered sets, or monotonic structures to optimize transitions.
What is the time complexity of Minimum Total Cost to Process All Elements?
The brute force solution takes exponential time because it explores every possible processing sequence. Optimized dynamic programming reduces this to O(n^2), while heap or prefix optimizations can improve it further to O(n log n) or O(n).