Date: 2026-06-06
Time: 18:00
This file solves LeetCode 1827 — Minimum Operations to Make the Array Increasing. It owns exactly one responsibility: given an integer array, compute the minimum number of increment-by-1 operations needed to make it strictly increasing.
min_operations(nums: list[int]) -> intThe sole public function. Contract:
nums[i] < nums[i+1].prev value instead of writing back into nums.Greedy single-pass. The algorithm makes the locally optimal choice at each position: if nums[i] isn't greater than prev, bump it to exactly prev + 1 — the smallest value that maintains strict increase. This greedy choice is globally optimal because making any element larger than necessary only forces more operations downstream.
Virtual state tracking. Rather than mutating the array, the function tracks prev as the "effective" value of the last element. This avoids side effects and keeps the function pure.
Imports: None — the function uses only built-in Python.
Imported by: The testsolution.py in its own directory. The massive "Imported By" list in the prompt is an artifact of the test harness structure (all test files likely share a common import pattern or runner), not actual usage of minoperations across hundreds of problems.
1. Initialize ops = 0 and prev = nums[0] — the first element never needs modification.
2. Iterate from index 1 through len(nums) - 1:
nums[i] <= prev: the element violates strict increase. Set prev += 1 (the minimum valid value), and accumulate prev - nums[i] into ops — that's the number of increments needed to bring nums[i] up to prev.nums[i] > prev: no operation needed. Update prev = nums[i].3. Return ops.
Example trace for [1, 5, 2, 4, 1]:
| i | nums[i] | prev (before) | action | ops |
|---|---------|---------------|--------|-----|
| 0 | 1 | 1 | init | 0 |
| 1 | 5 | 1 | 5 > 1, prev=5 | 0 |
| 2 | 2 | 5 | 2 <= 5, prev=6, ops += 4 | 4 |
| 3 | 4 | 6 | 4 <= 6, prev=7, ops += 3 | 7 |
| 4 | 1 | 7 | 1 <= 7, prev=8, ops += 7 | 14 |
i, prev is always strictly greater than the effective value at index i-1.prev is set to exactly prev_old + 1 when a violation occurs — never higher. This guarantees minimum total operations.nums[0] unconditionally on line prev = nums[0]. An empty list will raise IndexError.None. The function assumes valid input per LeetCode constraints (1 <= nums.length <= 5000). An empty list causes an unhandled IndexError. There is no validation, which is standard for competitive programming solutions.
minimum-operations-to-make-the-array-increasing/test_solution.py — See what edge cases the test suite covers (single-element, already sorted, all-equal, descending)minimum-operations-to-make-the-array-increasing/review.md — Code review notes that may capture alternative approaches or known limitationsremove-one-element-to-make-the-array-strictly-increasing/solution.py:canBeIncreasing — A related problem that asks whether removing one element suffices — different algorithmic structure for a similar invariantgreedy-vs-dp-for-array-ordering — When greedy single-pass works (like here) vs. when you need DP (e.g., longest increasing subsequence variants)min-ops-greedy-optimality — Setting prev = prev + 1 on violation is provably optimal; any larger target increases total ops without benefit to later elements.min-ops-no-mutation — The function never modifies the input list; it tracks effective values via the prev variable.min-ops-empty-input-crashes — Passing an empty list raises IndexError at prev = nums[0] — no guard exists.min-ops-linear-time — The algorithm runs in O(n) time and O(1) extra space, making a single pass over the array.