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3523. Make Array Non-decreasing

Description

You are given an integer array nums. In one operation, you can select a subarray and replace it with a single element equal to its maximum value.

Return the maximum possible size of the array after performing zero or more operations such that the resulting array is non-decreasing.

 

Example 1:

Input: nums = [4,2,5,3,5]

Output: 3

Explanation:

One way to achieve the maximum size is:

  1. Replace subarray nums[1..2] = [2, 5] with 5 → [4, 5, 3, 5].
  2. Replace subarray nums[2..3] = [3, 5] with 5 → [4, 5, 5].

The final array [4, 5, 5] is non-decreasing with size 3.

Example 2:

Input: nums = [1,2,3]

Output: 3

Explanation:

No operation is needed as the array [1,2,3] is already non-decreasing.

 

Constraints:

  • 1 <= nums.length <= 2 * 105
  • 1 <= nums[i] <= 2 * 105

Solutions

Solution 1

  • class Solution {
    public int maximumPossibleSize(int[] nums) {
        int ans = 0, mx = 0;
        for (int x : nums) {
            if (mx <= x) {
                ++ans;
                mx = x;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int maximumPossibleSize(vector<int>& nums) {
        int ans = 0, mx = 0;
        for (int x : nums) {
            if (mx <= x) {
                ++ans;
                mx = x;
            }
        }
        return ans;
    }
};

  • class Solution:
    def maximumPossibleSize(self, nums: List[int]) -> int:
        ans = mx = 0
        for x in nums:
            if mx <= x:
                ans += 1
                mx = x
        return ans


  • func maximumPossibleSize(nums []int) int {
	ans, mx := 0, 0
	for _, x := range nums {
		if mx <= x {
			ans++
			mx = x
		}
	}
	return ans
}

  • function maximumPossibleSize(nums: number[]): number {
    let [ans, mx] = [0, 0];
    for (const x of nums) {
        if (mx <= x) {
            ++ans;
            mx = x;
        }
    }
    return ans;
}

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