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LeetCode 2289: Steps to Make Array Non-decreasing Solution

Master LeetCode problem 2289 (Steps to Make Array Non-decreasing), a medium 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.

2289. Steps to Make Array Non-decreasing

Medium
ArrayLinked ListStackMonotonic Stack

Problem Explanation

Explanation:

To solve this problem, we need to iterate through the array and adjust elements to make the array non-decreasing. We can achieve this by modifying any element nums[i] such that nums[i - 1] <= nums[i]. If nums[i - 1] > nums[i], we need to adjust either nums[i - 1] or nums[i] to make the array non-decreasing.

We can count the number of steps needed to make the array non-decreasing by iterating through the array and adjusting elements whenever necessary. By doing this, we can keep track of the number of steps required.

Algorithm:

  1. Initialize a variable count to 0 to keep track of the number of steps.
  2. Iterate through the array starting from index 1.
  3. If nums[i - 1] > nums[i]:
    • Increment count.
    • If nums[i - 2] > nums[i], update nums[i] to nums[i - 1].
    • Otherwise, update nums[i - 1] to nums[i].
  4. After iterating through the array, return the final count.

Time Complexity: The time complexity of this algorithm is O(n) where n is the length of the input array nums.

Space Complexity: The space complexity of this algorithm is O(1) as we are not using any extra space other than a few variables.

:

Solution Code

class Solution {
    public int makeArrayNonDecreasing(int[] nums) {
        int count = 0;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i - 1] > nums[i]) {
                count++;
                if (i - 2 >= 0 && nums[i - 2] > nums[i]) {
                    nums[i] = nums[i - 1];
                } else {
                    nums[i - 1] = nums[i];
                }
            }
        }
        return count;
    }
}

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