Problem Description

You are given a target array target of positive integers. You can perform the following operation any number of times:

• Select any subarray (contiguous segment) of target and increment every element in that subarray by 1.

Your task is to determine the minimum number of such operations needed to transform an initial array of all zeros (of the same length as target) into the given target array.

Constraints:

• 1 ≤ target.length ≤ 10⁵
• 1 ≤ target[i] ≤ 10⁵

Thought Process

Let's start by thinking about the problem from a brute-force perspective. We could try to simulate every increment operation, but this would be far too slow for large arrays. Instead, we need to look for patterns or properties in the allowed operations.

Notice that when we increment a subarray, all elements in that range increase by one. To build up the target array from zeros, we need to "add" the required value to each position, but we want to do it as efficiently as possible, minimizing the number of operations.

If we increment the entire array, every element increases. But if only a part of the array needs to go up, we can focus on that segment. The key insight is to think about how much each position needs to be incremented compared to its previous position.

Rather than incrementing each individual element separately, we can take advantage of overlapping increments. This leads us to look for the minimal set of increments needed to reach the target, by examining the differences between consecutive elements.

Solution Approach

The optimal approach relies on the following observations:

• For the first element, we need to perform target[0] increments (since we start from zero).
• For each subsequent element target[i], if target[i] > target[i-1], we need target[i] - target[i-1] additional increments to reach its value (since previous increments would have already increased target[i] up to target[i-1]).
• If target[i] <= target[i-1], no new increments are needed for that element, as it has already been increased enough by earlier operations.

Therefore, the total number of operations required is:

• Sum of target[0] and the positive differences target[i] - target[i-1] for all i > 0.

Algorithm Steps:

1. Initialize operations = target[0].
2. Iterate from i = 1 to target.length - 1:
• If target[i] > target[i-1], add target[i] - target[i-1] to operations.
3. Return operations as the answer.

This approach is efficient, requiring only a single pass through the array.

Example Walkthrough

Let's work through an example with target = [1, 2, 3, 2, 1].

• Start: [0, 0, 0, 0, 0]
• To get to [1, 1, 1, 1, 1], increment the whole array once (1 operation).
• To get to [1, 2, 2, 2, 1], increment positions 1 to 3 once (2-1=1 operation).
• To get to [1, 2, 3, 2, 1], increment position 2 once (3-2=1 operation).
• The values then decrease, so no more increments are needed for positions 3 and 4.
• Total operations: 1 (for position 0) + 1 (for position 1) + 1 (for position 2) = 3

This matches our step-by-step calculation:

• Start: operations = 1
• i=1: target[1]=2, target[0]=1 ⇒ add 1 (total: 2)
• i=2: target[2]=3, target[1]=2 ⇒ add 1 (total: 3)
• i=3: target[3]=2, target[2]=3 ⇒ add 0
• i=4: target[4]=1, target[3]=2 ⇒ add 0

Time and Space Complexity

Brute-force Approach:

• Simulating every increment for every subarray would require O(n²) or worse, which is infeasible for large n.

• We only need a single pass through the array.
• Time Complexity: O(n), where n is the length of target.
• Space Complexity: O(1), since we only use a few variables for counting.

This efficiency makes the solution practical for large input sizes.

Code Implementation

class Solution:
    def minNumberOperations(self, target):
        operations = target[0]
        for i in range(1, len(target)):
            if target[i] > target[i-1]:
                operations += target[i] - target[i-1]
        return operations
      

class Solution {
public:
    int minNumberOperations(vector<int>& target) {
        int operations = target[0];
        for (int i = 1; i < target.size(); ++i) {
            if (target[i] > target[i-1]) {
                operations += target[i] - target[i-1];
            }
        }
        return operations;
    }
};
      

class Solution {
    public int minNumberOperations(int[] target) {
        int operations = target[0];
        for (int i = 1; i < target.length; i++) {
            if (target[i] > target[i - 1]) {
                operations += target[i] - target[i - 1];
            }
        }
        return operations;
    }
}
      

var minNumberOperations = function(target) {
    let operations = target[0];
    for (let i = 1; i < target.length; i++) {
        if (target[i] > target[i - 1]) {
            operations += target[i] - target[i - 1];
        }
    }
    return operations;
};
      

Summary

The core insight of this problem is to recognize that we only need to count the increments required at each "uphill" (increase) in the target array, starting from zero. By summing the first value and all increases over the previous value, we efficiently compute the minimum number of required subarray increment operations. This approach avoids unnecessary simulation and leverages the structure of the problem for a clean and optimal solution.