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Minimum Operations to Make the Array Increasing

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ArraysGreedy Algorithms

You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1.

For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,3,3].

Return the minimum number of operations needed to make nums strictly increasing.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Example 1:

Input: nums = [1,1,1]
Output: 3
Explanation: You can do the following operations:
1) Increment nums[2], so nums becomes [1,1,2].
2) Increment nums[1], so nums becomes [1,2,2].
3) Increment nums[2], so nums becomes [1,2,3].

Example 2:

Input: nums = [1,5,2,4,1]
Output: 14

Example 3:

Input: nums = [8]
Output: 0

Constraints:

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

Try coding it in LeetCode

Minimum Operations to Make the Array Increasing

Easy

Asked by:

14 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1.

For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,3,3].

Return the minimum number of operations needed to make nums strictly increasing.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Example 1:

Input: nums = [1,1,1]
Output: 3
Explanation: You can do the following operations:
1) Increment nums[2], so nums becomes [1,1,2].
2) Increment nums[1], so nums becomes [1,2,2].
3) Increment nums[2], so nums becomes [1,2,3].

Example 2:

Input: nums = [1,5,2,4,1]
Output: 14

Example 3:

Input: nums = [8]
Output: 0

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers within the array?
Is the input array guaranteed to be non-empty, or should I handle the case where the array is null or empty?
If the array is already increasing, should I return 0?
Are all elements integers or could floating-point numbers be present?
Can I modify the input array directly, or should I avoid making changes?

Brute Force Solution

Approach

The brute force way to solve this is to try every single possible way to make the numbers increase one by one. We modify the numbers and check if the sequence is increasing after each change, counting our modifications.

Here's how the algorithm would work step-by-step:

Start with the first two numbers.
If the second number is not bigger than the first, increase it by one. Count this as an operation.
Now look at the second and third numbers. If the third isn't bigger than the second, increase it by one and count that operation.
Continue doing this for all consecutive pairs of numbers. Every time you make a number bigger, remember to count the operation.
After you've gone through the whole list, check if all the numbers are now in increasing order.
If they are, that is one possible solution, and we'll keep track of the number of operations it took.
Since this is brute force, we will explore other ways to increase the numbers and record the number of operations each way takes.
After exploring all the possibilities, we will pick the method that requires the fewest operations and return that value.

Code Implementation

def min_operations_brute_force(numbers):
    number_of_operations = 0
    modified_numbers = numbers[:]

    # Iterate through the list, ensuring each element is greater than the last
    for index in range(len(modified_numbers) - 1):
        #If the current number is not bigger than the previous number, increase it
        if modified_numbers[index + 1] <= modified_numbers[index]:
            difference = modified_numbers[index] - modified_numbers[index + 1] + 1
            modified_numbers[index + 1] += difference
            number_of_operations += difference

    #Check if the array is increasing. If not, return -1.
    for index in range(len(modified_numbers) - 1):
        if modified_numbers[index + 1] <= modified_numbers[index]:
            return -1

    return number_of_operations

Big(O) Analysis

Time Complexity

O(n^n) – The brute force approach outlined explores all possible ways to increase numbers to achieve a strictly increasing sequence. For each element, there could be multiple choices of how much to increase it. In the worst-case scenario, we might have to consider increasing each of the n elements up to n different values to explore all combinations. This generates a decision tree where each node has up to n children, and the depth of the tree is n. Therefore, the number of possible paths, and thus the number of operations explored, grows exponentially, approaching O(n^n).

Space Complexity

O(1) – The brute force solution described explores different ways to increase the array elements. However, based on the plain English description, it seems to modify the input array in place and does not create significant auxiliary data structures. It implicitly keeps track of the minimum number of operations but it does so by updating a single variable. No temporary lists, hash maps, or recursion are mentioned that would grow with the input size N (the length of the array), therefore the auxiliary space used is constant.

Optimal Solution

Approach

The goal is to make each number in a sequence bigger than the one before it. To do this most efficiently, we check each number and increase it only as much as needed to meet the 'bigger than the previous' rule, keeping the total increase as small as possible.

Here's how the algorithm would work step-by-step:

Start by looking at the second number in the sequence.
Compare it to the number before it.
If the second number isn't bigger, calculate how much you need to increase it to be just one greater than the previous number.
Record this increase as an 'operation'.
Update the second number to its new, increased value.
Move to the next number and repeat the comparison and increasing process with its previous number.
Keep doing this until you've checked and adjusted all the numbers in the sequence.
The total number of 'operations' you recorded is the minimum number of increases needed to make the sequence steadily increasing.

Code Implementation

def min_operations_to_make_array_increasing(nums):
    operation_count = 0

    for i in range(1, len(nums)):
        # Compare the current element with the previous one.
        if nums[i] <= nums[i - 1]:
            # Calculate the increase needed
            increase_needed = nums[i - 1] - nums[i] + 1

            # Increment operation counter.
            operation_count += increase_needed

            # Update current number to make it greater than previous
            nums[i] += increase_needed

    return operation_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array once, where n is the size of the input array. Inside the loop, it performs a constant number of operations: a comparison and, potentially, an addition and an assignment. Thus, the time complexity is directly proportional to the number of elements in the array, resulting in O(n) time complexity.

Space Complexity

O(1) – The provided algorithm iterates through the input array, modifying it in place. It only uses a single variable to keep track of the number of operations, and another to access the elements during iteration. The space required does not scale with the size of the input array (N), therefore the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no operations are needed on an empty or null array to make it increasing.
Array with a single element	Return 0 since a single-element array is already considered increasing.
Array with two elements already in increasing order	Return 0 since no operations are needed if the array is already increasing.
Array with two elements in decreasing order	Return the absolute difference between the second and first element plus 1, since only one operation is needed.
Array with elements having large positive or negative values, potentially leading to integer overflow during increment operations	Use a data type that supports larger values (e.g., long) to prevent integer overflow during increment operations.
Array with elements all equal to the same value	Increment each subsequent element to make it one greater than the previous one, accumulating the total operations.
Very large array size impacting performance	The solution iterates through the array once, so it should scale linearly (O(n)) and remain efficient even for large arrays.
Array containing both positive and negative numbers including zero	The algorithm should correctly handle these values, incrementing elements as needed to ensure an increasing sequence.

Null or empty input array

How to Handle:

Return 0 since no operations are needed on an empty or null array to make it increasing.

Array with a single element

How to Handle:

Return 0 since a single-element array is already considered increasing.

Array with two elements already in increasing order

How to Handle:

Return 0 since no operations are needed if the array is already increasing.

Array with two elements in decreasing order

How to Handle:

Return the absolute difference between the second and first element plus 1, since only one operation is needed.

Array with elements having large positive or negative values, potentially leading to integer overflow during increment operations

How to Handle:

Use a data type that supports larger values (e.g., long) to prevent integer overflow during increment operations.

Array with elements all equal to the same value

How to Handle:

Increment each subsequent element to make it one greater than the previous one, accumulating the total operations.

Very large array size impacting performance

How to Handle:

The solution iterates through the array once, so it should scale linearly (O(n)) and remain efficient even for large arrays.

Array containing both positive and negative numbers including zero

How to Handle:

The algorithm should correctly handle these values, incrementing elements as needed to ensure an increasing sequence.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers within the array?
Is the input array guaranteed to be non-empty, or should I handle the case where the array is null or empty?
If the array is already increasing, should I return 0?
Are all elements integers or could floating-point numbers be present?
Can I modify the input array directly, or should I avoid making changes?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the first two numbers.
If the second number is not bigger than the first, increase it by one. Count this as an operation.
Now look at the second and third numbers. If the third isn't bigger than the second, increase it by one and count that operation.
Continue doing this for all consecutive pairs of numbers. Every time you make a number bigger, remember to count the operation.
After you've gone through the whole list, check if all the numbers are now in increasing order.
If they are, that is one possible solution, and we'll keep track of the number of operations it took.
Since this is brute force, we will explore other ways to increase the numbers and record the number of operations each way takes.
After exploring all the possibilities, we will pick the method that requires the fewest operations and return that value.

Code Implementation

def min_operations_brute_force(numbers):
    number_of_operations = 0
    modified_numbers = numbers[:]

    # Iterate through the list, ensuring each element is greater than the last
    for index in range(len(modified_numbers) - 1):
        #If the current number is not bigger than the previous number, increase it
        if modified_numbers[index + 1] <= modified_numbers[index]:
            difference = modified_numbers[index] - modified_numbers[index + 1] + 1
            modified_numbers[index + 1] += difference
            number_of_operations += difference

    #Check if the array is increasing. If not, return -1.
    for index in range(len(modified_numbers) - 1):
        if modified_numbers[index + 1] <= modified_numbers[index]:
            return -1

    return number_of_operations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by looking at the second number in the sequence.
Compare it to the number before it.
If the second number isn't bigger, calculate how much you need to increase it to be just one greater than the previous number.
Record this increase as an 'operation'.
Update the second number to its new, increased value.
Move to the next number and repeat the comparison and increasing process with its previous number.
Keep doing this until you've checked and adjusted all the numbers in the sequence.
The total number of 'operations' you recorded is the minimum number of increases needed to make the sequence steadily increasing.

Code Implementation

def min_operations_to_make_array_increasing(nums):
    operation_count = 0

    for i in range(1, len(nums)):
        # Compare the current element with the previous one.
        if nums[i] <= nums[i - 1]:
            # Calculate the increase needed
            increase_needed = nums[i - 1] - nums[i] + 1

            # Increment operation counter.
            operation_count += increase_needed

            # Update current number to make it greater than previous
            nums[i] += increase_needed

    return operation_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no operations are needed on an empty or null array to make it increasing.
Array with a single element	Return 0 since a single-element array is already considered increasing.
Array with two elements already in increasing order	Return 0 since no operations are needed if the array is already increasing.
Array with two elements in decreasing order	Return the absolute difference between the second and first element plus 1, since only one operation is needed.
Array with elements having large positive or negative values, potentially leading to integer overflow during increment operations	Use a data type that supports larger values (e.g., long) to prevent integer overflow during increment operations.
Array with elements all equal to the same value	Increment each subsequent element to make it one greater than the previous one, accumulating the total operations.
Very large array size impacting performance	The solution iterates through the array once, so it should scale linearly (O(n)) and remain efficient even for large arrays.
Array containing both positive and negative numbers including zero	The algorithm should correctly handle these values, incrementing elements as needed to ensure an increasing sequence.

Null or empty input array

How to Handle:

Return 0 since no operations are needed on an empty or null array to make it increasing.

Array with a single element

How to Handle:

Return 0 since a single-element array is already considered increasing.

Array with two elements already in increasing order

How to Handle:

Return 0 since no operations are needed if the array is already increasing.

Array with two elements in decreasing order

How to Handle:

Return the absolute difference between the second and first element plus 1, since only one operation is needed.

Array with elements having large positive or negative values, potentially leading to integer overflow during increment operations

How to Handle:

Use a data type that supports larger values (e.g., long) to prevent integer overflow during increment operations.

Array with elements all equal to the same value

How to Handle:

Increment each subsequent element to make it one greater than the previous one, accumulating the total operations.

Very large array size impacting performance

How to Handle:

The solution iterates through the array once, so it should scale linearly (O(n)) and remain efficient even for large arrays.

Array containing both positive and negative numbers including zero

How to Handle:

The algorithm should correctly handle these values, incrementing elements as needed to ensure an increasing sequence.