1827. Minimum Operations to Make the Array Increasing

Problem Description

The problem provides us with an array of integers called nums, which uses 0-based indexing. We are tasked with finding the minimum number of operations needed to make this array strictly increasing. An operation consists of selecting any element in the array and incrementing it by 1.

A strictly increasing array is defined as one in which every element is less than the element immediately following it (nums[i] < nums[i+1]). A single-element array is considered strictly increasing by default since there are no adjacent elements to compare.

The ultimate goal here is to ensure that for every pair of adjacent elements (nums[i], nums[i+1]), the condition nums[i] < nums[i+1] holds true, by performing the least number of increment operations.

Intuition

The intuition behind the solution involves ensuring that for each element nums[i] in the array, if it’s not already greater than the previous element nums[i-1], we increment it enough times such that it becomes 1 more than the previous element.

To achieve this, we track the maximum value (mx) needed so far as we iterate through the array. For each value v in the array, if v is smaller or equal to mx, we know we need to increment v to at least mx + 1 to maintain the strictly increasing property.

We calculate any gap that might exist between mx + 1 (which v needs to be to keep the array strictly increasing) and the current value v. This gap represents the number of increments needed for the current element v. We add this gap to our running total of operations (ans). After considering v, we update mx to be the maximum of mx + 1 (to ensure strict increasing order) and v (in case the current value is already large enough and doesn't need increments).

Here's the flow:

We initialize the ans (answer) variable to track the total number of operations performed and mx (maximum needed so far) with the value 0.
We iterate through each value v in nums.
If v is less than or equal to mx, we calculate the difference mx + 1 - v (the number of operations needed to make the current element strictly larger than the previous one) and add it to ans. If v is already greater than mx, no operations are needed, so we would add 0.
We update mx to be the greater value between mx + 1 and v to ensure that we're always setting mx to be at least one greater than the last value (to maintain the strictly increasing order) or to account for the current value if it doesn't need to be incremented.
After going through all elements in nums, the value of ans will be the minimum number of operations required to make nums strictly increasing.

Learn more about Greedy patterns.

Solution Approach

The solution uses a simple, yet efficient algorithm to resolve the challenge. It does not require complex data structures or patterns. The straightforward use of a for-loop and basic arithmetic operations in combination with simple variable tracking proves to be efficient for this problem.

Here's a detailed walkthrough of the implementation based on the reference solution:

We start by initializing two variables, ans and mx, to 0. ans will keep track of the total number of operations performed, while mx will hold the current maximum value needed to maintain a strictly increasing sequence.
The core part of our solution is a for-loop that iterates through each number v in the input array nums. This loop is where we determine if an increment operation is necessary and if so, how many:
```
1for v in nums:
2    ans += max(0, mx + 1 - v)
3    mx = max(mx + 1, v)
```
Let's break down the loop operations:
- ans += max(0, mx + 1 - v): We calculate the difference between mx + 1 and v, which gives us the number of operations needed to make the current number v comply with the strictly increasing criterion. We use max(0, mx + 1 - v) because if v is already larger than mx, we do not need to perform any operations, hence we add zero to ans.
- mx = max(mx + 1, v): We update mx to be the larger of mx + 1 and v. This operation is crucial because it ensures that we will always compare subsequent numbers to a value that keeps the sequence strictly increasing. If v is already equal to or larger than mx + 1, we set mx to v. Otherwise, we ensure mx becomes mx + 1, which is the minimum value the next number in the sequence must exceed.
Once the loop completes, the variable ans will contain the sum of all the increments performed, which is the total number of operations needed to make the array nums strictly increasing. This value is then returned as the solution.

The simplicity of the algorithm comes from the realization that we can keep the problem state using only two variables and do not need to modify the original array. Essentially, it's the concept of dynamic programming without the need for memoization or auxiliary data structures, as we only care about the relationship between adjacent elements. The time complexity of the algorithm is O(n), where n is the number of elements in nums, because we only need to iterate through the array once. The space complexity is O(1) because we use a constant amount of extra space.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we have the following array:

1nums = [3, 4, 2, 6, 1]

We want to perform the minimum number of operations to make this array strictly increasing.

Let's apply our algorithm:

Initialize ans and mx to 0. These will keep track of the total number of operations and the maximum value needed respectively.
Start the for-loop with the first element v = 3. Since mx is 0, mx + 1 - v is -2. We don't need to perform any increments because v is already greater than mx. Update ans = 0 and mx = max(1, 3) = 3.
Next, v = 4. No increments needed, as v is greater than mx. Update ans = 0 and mx = max(4, 4) = 4.
Now, v = 2. Since 2 is not greater than mx, we need to increment v by mx + 1 - v = 4 + 1 - 2 = 3 times. Update ans = 0 + 3 = 3 and now mx = max(4 + 1, 2) = 5.
Then, v = 6. No increments needed, as v is greater than mx. Update ans = 3 and mx = max(5 + 1, 6) = 6.
Lastly, v = 1. v is less than mx, we must increment v by mx + 1 - v = 6 + 1 - 1 = 6 times. Update ans = 3 + 6 = 9 and finally mx = max(6 + 1, 1) = 7.

After going through each element, we found the total number of operations required to make nums strictly increasing is ans = 9.

In conclusion, the output for the array nums = [3, 4, 2, 6, 1] would be 9, meaning we need to perform 9 increment operations to transform it into a strictly increasing array.

Solution Implementation

1class Solution:
2    def minOperations(self, nums: List[int]) -> int:
3        # Initialize the answer counter to count the minimum operations required
4        operations_count = 0
5      
6        # Initialize the max_value variable to keep track of the maximum integer seen so far
7        max_value = 0
8      
9        # Loop through each value in the given list
10        for value in nums:
11            # If the current value is less than or equal to the max_value adjusted by one, 
12            # calculate the operations needed to make it one greater than the max_value seen so far
13            operations_count += max(0, max_value + 1 - value)
14          
15            # Update max_value: it should be the maximum of the previous max_value adjusted by one, 
16            # or the current value in the list in case it's larger
17            max_value = max(max_value + 1, value)
18      
19        # Return the total number of operations counted
20        return operations_count
21

1class Solution {
2    public int minOperations(int[] nums) {
3        int operations = 0; // To store the minimum number of operations required
4        int maxVal = 0; // To keep track of the maximum value obtained so far
5      
6        // Iterate through all elements in the array
7        for (int value : nums) {
8            // If the current value is less than the maxVal + 1,
9            // we need to increment it, which counts as an operation
10            operations += Math.max(0, maxVal + 1 - value); // Add necessary operations
11          
12            // Update the maxVal to be the maximum of the current value and maxVal + 1,
13            // since we want to ensure the next number is at least maxVal + 1
14            maxVal = Math.max(maxVal + 1, value); // Update the maxVal
15        }
16      
17        return operations; // Return the total number of operations
18    }
19}
20

1#include <vector>
2#include <algorithm>
3
4class Solution {
5public:
6    // Function to calculate the minimum number of operations needed
7    // to make the array strictly increasing
8    int minOperations(std::vector<int>& nums) {
9        int totalOperations = 0; // Variable to keep track of total operations performed
10        int maxSoFar = 0; // Variable to keep track of the maximum value encountered so far
11      
12        // Loop through each element in the vector
13        for (int& value : nums) {
14            // Calculate operations needed for current element to be greater
15            // than the maxSoFar. If the value is already greater than maxSoFar,
16            // no operations are needed; hence, we use max with 0.
17            totalOperations += std::max(0, maxSoFar + 1 - value);
18          
19            // Update maxSoFar to be either the current value or one more
20            // than maxSoFar, whichever is larger, to maintain strict increasing order.
21            maxSoFar = std::max(maxSoFar + 1, value);
22        }
23      
24        return totalOperations; // Return the total number of operations
25    }
26};
27

1/**
2 * Calculates the minimum number of operations needed to make the array strictly increasing.
3 * Each operation consists of incrementing a number in the array.
4 * @param {number[]} nums - The input array of numbers.
5 * @returns {number} The minimum number of operations required.
6 */
7function minOperations(nums: number[]): number {
8    // Initialize the number of operations (ans) to 0
9    let operationsCount = 0;
10
11    // Initialize the maximum number seen so far to enable comparisons
12    let currentMax = 0;
13
14    // Iterate through each number in the input array
15    for (const value of nums) {
16        // Calculate the number of operations needed for the current number, if any,
17        // ensuring the number is at least one more than the current maximum
18        operationsCount += Math.max(0, currentMax + 1 - value);
19
20        // Update the current maximum to be either the incremented maximum or the current value,
21        // whichever is larger, to maintain the strictly increasing property
22        currentMax = Math.max(currentMax + 1, value);
23    }
24
25    // Return the total number of operations needed
26    return operationsCount;
27}
28

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n), where n is the length of the nums array. This is because there is a single for-loop that iterates over all elements of the array once, performing a constant number of operations for each element. The operations performed within the loop (calculations and comparisons) are all constant time operations.

Space Complexity

The space complexity of the given code is O(1) (constant space). This is because the amount of extra space used does not depend on the input size n. The code only uses a fixed number of variables (ans, mx, and v) that do not expand with the size of the input array.

Learn more about how to find time and space complexity quickly using problem constraints.

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Question 1 out of 10

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12

1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18

1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

1 2 3 6 x x

1 2 3 x x x 6 x x

1 2 6 x x x 3 x x

I don’t know