#1827

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Operations to Make the Array Increasing

Calculate the minimum number of increments required to transform a given integer array into a strictly increasing sequence efficiently.

array greedy

Problem Statement

Given an integer array nums, you can increment any element by 1 in a single operation. Determine the fewest total increments required to make the array strictly increasing.

An array is strictly increasing if each element is less than the next. Return the total minimum operations needed. Single-element arrays are trivially strictly increasing.

Examples

Example 1

Input: nums = [1,1,1]

Output: 3

You can do the following operations:

Increment nums[2], so nums becomes [1,1,2].
Increment nums[1], so nums becomes [1,2,2].
Increment nums[2], so nums becomes [1,2,3].

Example 2

Input: nums = [1,5,2,4,1]

Output: 14

Example details omitted.

Example 3

Input: nums = [8]

Output: 0

Example details omitted.

Constraints

1 <= nums.length <= 5000
1 <= nums[i] <= 104

Solution Approach

Greedy Increment Traversal

Iterate from left to right. For each nums[i+1], if it is less than or equal to nums[i], increment nums[i+1] to nums[i]+1 and add the difference to the operation count. This directly implements the greedy choice plus invariant validation pattern.

In-place Array Update

Update the array in-place during traversal to maintain the strictly increasing invariant. This avoids extra space while ensuring each decision accounts for previous increments, preventing undercounting operations.

Summation of Operations

Keep a running total of all increments performed. This total represents the minimum number of operations required. The correctness relies on always satisfying nums[i+1] >= nums[i]+1 at each step.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since each element is visited once. Space complexity is O(1) extra if modifying in-place, otherwise O(n) if using a separate array to track increments.

What Interviewers Usually Probe

Notice the array must be strictly increasing, hinting at element-wise comparison.
Greedy approach is expected, incrementing only when necessary to maintain the invariant.
Tracking total operations while scanning sequentially shows awareness of minimal adjustments.

Common Pitfalls or Variants

Common pitfalls

Failing to compare nums[i+1] with the already updated nums[i], causing undercounted operations.
Using nested loops instead of a single pass, resulting in unnecessary time complexity.
Ignoring the edge case where the array has only one element, which is trivially strictly increasing.

Follow-up variants

Compute minimum decrements to make the array strictly decreasing using a similar greedy pattern.
Find minimum operations if allowed to increment or decrement each element by 1, balancing adjustments.
Apply the approach to 2D arrays, ensuring each row or column is strictly increasing.

FAQ

What is the key pattern to solve Minimum Operations to Make the Array Increasing?

The key pattern is greedy choice plus invariant validation: always increment the next element to exceed the previous by at least 1.

Can I solve this problem without modifying the original array?

Yes, by tracking the required increments in a separate variable while comparing elements, preserving the original array.

Why is a single traversal enough for this problem?

Because the greedy choice ensures each element satisfies the strictly increasing condition once, preventing the need for revisiting elements.

How do I handle an array of length 1?

Single-element arrays are already strictly increasing, so the minimum number of operations is zero.

What common mistake should I avoid in this greedy approach?

Do not forget to compare nums[i+1] with the updated nums[i] after previous increments, otherwise the operation count will be incorrect.

terminal

Solution

Solution 1: Single Pass

We use a variable $mx$ to record the maximum value of the current strictly increasing array, initially $mx = 0$.

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = mx = 0
        for v in nums:
            ans += max(0, mx + 1 - v)
            mx = max(mx + 1, v)
        return ans

Continue Practicing

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