LeetCode 1827 - Minimum Operations to Make the Array Increasing

LeetCode Problem 1827

Difficulty: 🟢 Easy
Topics: Array, Greedy

Solution

Problem Understanding

The problem asks us to transform a given integer array nums into a strictly increasing array using the minimum number of operations. An operation consists of incrementing any element of the array by 1. A strictly increasing array is one where each element is strictly less than the next element, i.e., nums[i] < nums[i+1] for all valid indices i.

The input is an array of integers, ranging from length 1 to 5000, with each integer between 1 and 10,000. The output is a single integer representing the minimum number of increment operations required to satisfy the strictly increasing condition.

Important edge cases include arrays that are already strictly increasing, arrays with all elements the same, arrays of length 1, and arrays with large jumps or repeated elements. These edge cases ensure the algorithm correctly handles minimal operations, avoids unnecessary increments, and does not produce errors with small or large inputs.

Approaches

The brute-force approach is straightforward: iterate through the array from left to right, and for each element nums[i], check if it is greater than the previous element nums[i-1]. If not, increment it one by one until it satisfies nums[i] > nums[i-1]. While this approach is correct, incrementing each element one by one can be very slow for large arrays or large numbers, potentially causing O(n * max(nums)) time complexity, which is inefficient for the given constraints.

The optimal approach leverages the observation that if nums[i] <= nums[i-1], the minimum number of increments needed is exactly nums[i-1] - nums[i] + 1. By applying this formula, we can directly compute the required operations in a single pass through the array without looping to increment each element individually. This reduces the time complexity to O(n) and only requires a constant amount of extra space.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n * max(nums))	O(1)	Increment elements one by one until strictly increasing
Optimal	O(n)	O(1)	Compute exact increment needed per element in one pass

Algorithm Walkthrough

1. Initialize a variable operations to 0 to count the total number of increments performed.
2. Iterate over the array starting from index 1 to the end.
3. For each element nums[i], compare it with the previous element nums[i-1].
4. If nums[i] <= nums[i-1], calculate the required increment as needed = nums[i-1] - nums[i] + 1.
5. Add needed to the operations counter.
6. Update nums[i] by adding needed to ensure the array remains strictly increasing.
7. Continue to the next element until the end of the array.
8. Return the total operations as the result.

Why it works: The algorithm guarantees a strictly increasing array because we always ensure nums[i] > nums[i-1] by applying the minimal necessary increment. By performing this check sequentially from left to right, no future operation can violate the previously established ordering, making the solution correct.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        operations = 0
        for i in range(1, len(nums)):
            if nums[i] <= nums[i - 1]:
                needed = nums[i - 1] - nums[i] + 1
                operations += needed
                nums[i] += needed
        return operations

The Python implementation initializes the operations counter, iterates through the array, checks if the current element is less than or equal to the previous element, and updates the element with the exact number of increments needed. Each increment is accumulated into operations, which is returned at the end.

Go Solution

func minOperations(nums []int) int {
    operations := 0
    for i := 1; i < len(nums); i++ {
        if nums[i] <= nums[i-1] {
            needed := nums[i-1] - nums[i] + 1
            operations += needed
            nums[i] += needed
        }
    }
    return operations
}

The Go version mirrors the Python solution. It uses a for loop starting from index 1, calculates the required increment when necessary, updates the array, and accumulates the operations count. Since slices are mutable in Go, in-place updates work similarly to Python.

Worked Examples

Example 1: nums = [1,1,1]

i	nums[i-1]	nums[i]	needed	operations	nums after update
1	1	1	1	1	[1,2,1]
2	2	1	2	3	[1,2,3]

Output: 3

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

i	nums[i-1]	nums[i]	needed	operations	nums after update
1	1	5	0	0	[1,5,2,4,1]
2	5	2	4	4	[1,5,6,4,1]
3	6	4	3	7	[1,5,6,7,1]
4	7	1	7	14	[1,5,6,7,8]

Output: 14

Example 3: nums = [8]

No operations are needed for a single-element array.

Output: 0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single pass through the array, performing constant time operations per element
Space	O(1)	Only a few integer variables are used, no extra space proportional to input

The complexity is optimal for the given constraints because the algorithm touches each element exactly once and uses no additional data structures.

Test Cases

# Provided examples
assert Solution().minOperations([1,1,1]) == 3  # All elements equal
assert Solution().minOperations([1,5,2,4,1]) == 14  # Mixed increases and decreases
assert Solution().minOperations([8]) == 0  # Single element

# Edge and boundary cases
assert Solution().minOperations([1,2,3,4,5]) == 0  # Already strictly increasing
assert Solution().minOperations([5,4,3,2,1]) == 10  # Strictly decreasing
assert Solution().minOperations([1,1,2,2,3,3]) == 6  # Repeated elements
assert Solution().minOperations([10000]*5000) == 12497500  # Maximum size and value

Test	Why
[1,1,1]	Handles repeated elements needing multiple increments
[1,5,2,4,1]	Handles alternating increases and decreases
[8]	Single-element edge case
[1,2,3,4,5]	Already strictly increasing array
[5,4,3,2,1]	Strictly decreasing array requiring maximum operations
[1,1,2,2,3,3]	Repeated elements scattered through array
[10000]*5000	Stress test for maximum input size and values

Edge Cases

Single-element array: A single-element array is trivially strictly increasing. The algorithm correctly returns 0 because the for loop does not execute.

Repeated elements: Arrays with repeated numbers require careful calculation of increments. The formula nums[i-1] - nums[i] + 1 ensures the minimum necessary increment to maintain strict increase.

Large arrays with large numbers: With arrays up to length 5000 and values up to 10,000, iterating efficiently is crucial. The optimal O(n) solution handles these limits without looping unnecessarily or causing integer overflow. In Python, integers are arbitrary precision; in Go, the calculations fit within the 32-bit range given constraints, but in practice using int is safe for these limits.