LeetCode 1526: Minimum Number of Increments on Subarrays to Form a Target Array Solution

Master LeetCode problem 1526 (Minimum Number of Increments on Subarrays to Form a Target Array), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1526. Minimum Number of Increments on Subarrays to Form a Target Array

Hard

Array Dynamic Programming Stack Greedy Monotonic Stack

Problem Explanation

Explanation:

To solve this problem, we can iterate through the target array and maintain a variable prev to keep track of the maximum value encountered so far. For each element in the target array, we calculate the difference between the current element and the previous element. If this difference is positive, it means we need to increment the subarray from the previous element to the current element to form the target array.

We add this difference to the total number of operations needed and update the prev variable. Finally, we return the total number of operations as the result.

Algorithm:

Initialize prev as 0 and operations as 0.
Iterate through the target array.
- For each element num:
  - Calculate the difference diff between num and prev.
  - If diff is positive, increment operations by diff.
  - Update prev to num.
Return operations as the result.

Time Complexity:

The time complexity of this approach is O(N), where N is the length of the target array.

Space Complexity:

The space complexity of this approach is O(1) as we are using only a constant amount of extra space.

: :

Solution Code

class Solution {
    public int minNumberOperations(int[] target) {
        int prev = 0;
        int operations = 0;

        for (int num : target) {
            int diff = num - prev;
            if (diff > 0) {
                operations += diff;
            }
            prev = num;
        }

        return operations;
    }
}

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