2771. Longest Non-decreasing Subarray From Two Arrays
Description
You are given two 0-indexed integer arrays nums1 and nums2 of length n.
Let's define another 0-indexed integer array, nums3, of length n. For each index i in the range [0, n - 1], you can assign either nums1[i] or nums2[i] to nums3[i].
Your task is to maximize the length of the longest non-decreasing subarray in nums3 by choosing its values optimally.
Return an integer representing the length of the longest non-decreasing subarray in nums3.
Note: A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums1 = [2,3,1], nums2 = [1,2,1] Output: 2 Explanation: One way to construct nums3 is: nums3 = [nums1[0], nums2[1], nums2[2]] => [2,2,1]. The subarray starting from index 0 and ending at index 1, [2,2], forms a non-decreasing subarray of length 2. We can show that 2 is the maximum achievable length.
Example 2:
Input: nums1 = [1,3,2,1], nums2 = [2,2,3,4] Output: 4 Explanation: One way to construct nums3 is: nums3 = [nums1[0], nums2[1], nums2[2], nums2[3]] => [1,2,3,4]. The entire array forms a non-decreasing subarray of length 4, making it the maximum achievable length.
Example 3:
Input: nums1 = [1,1], nums2 = [2,2] Output: 2 Explanation: One way to construct nums3 is: nums3 = [nums1[0], nums1[1]] => [1,1]. The entire array forms a non-decreasing subarray of length 2, making it the maximum achievable length.
Constraints:
1 <= nums1.length == nums2.length == n <= 1051 <= nums1[i], nums2[i] <= 109
Solutions
Solution 1: Dynamic Programming
We define two variables $f$ and $g$, which represent the length of the longest non-decreasing subarray at the current position. Here, $f$ represents the length of the longest non-decreasing subarray ending with an element from $nums1$, and $g$ represents the length of the longest non-decreasing subarray ending with an element from $nums2$. Initially, $f = g = 1$, and the initial answer $ans = 1$.
Next, we iterate over the array elements in the range $i \in [1, n)$, and for each $i$, we define two variables $ff$ and $gg$, which represent the length of the longest non-decreasing subarray ending with $nums1[i]$ and $nums2[i]$ respectively. When initialized, $ff = gg = 1$.
We can calculate the values of $ff$ and $gg$ based on the values of $f$ and $g$:
If $nums1[i] \ge nums1[i - 1]$, then $ff = \max(ff, f + 1)$;
If $nums1[i] \ge nums2[i - 1]$, then $ff = \max(ff, g + 1)$;
If $nums2[i] \ge nums1[i - 1]$, then $gg = \max(gg, f + 1)$;
If $nums2[i] \ge nums2[i - 1]$, then $gg = \max(gg, g + 1)$.
Then, we update $f = ff$ and $g = gg$, and update $ans$ to $\max(ans, f, g)$.
After the iteration ends, we return $ans$.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
Python3
class Solution:
def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
n = len(nums1)
f = g = 1
ans = 1
for i in range(1, n):
ff = gg = 1
if nums1[i] >= nums1[i - 1]:
ff = max(ff, f + 1)
if nums1[i] >= nums2[i - 1]:
ff = max(ff, g + 1)
if nums2[i] >= nums1[i - 1]:
gg = max(gg, f + 1)
if nums2[i] >= nums2[i - 1]:
gg = max(gg, g + 1)
f, g = ff, gg
ans = max(ans, f, g)
return ans
Java
class Solution {
public int maxNonDecreasingLength(int[] nums1, int[] nums2) {
int n = nums1.length;
int f = 1, g = 1;
int ans = 1;
for (int i = 1; i < n; ++i) {
int ff = 1, gg = 1;
if (nums1[i] >= nums1[i - 1]) {
ff = Math.max(ff, f + 1);
}
if (nums1[i] >= nums2[i - 1]) {
ff = Math.max(ff, g + 1);
}
if (nums2[i] >= nums1[i - 1]) {
gg = Math.max(gg, f + 1);
}
if (nums2[i] >= nums2[i - 1]) {
gg = Math.max(gg, g + 1);
}
f = ff;
g = gg;
ans = Math.max(ans, Math.max(f, g));
}
return ans;
}
}
C++
class Solution {
public:
int maxNonDecreasingLength(vector<int>& nums1, vector<int>& nums2) {
int n = nums1.size();
int f = 1, g = 1;
int ans = 1;
for (int i = 1; i < n; ++i) {
int ff = 1, gg = 1;
if (nums1[i] >= nums1[i - 1]) {
ff = max(ff, f + 1);
}
if (nums1[i] >= nums2[i - 1]) {
ff = max(ff, g + 1);
}
if (nums2[i] >= nums1[i - 1]) {
gg = max(gg, f + 1);
}
if (nums2[i] >= nums2[i - 1]) {
gg = max(gg, g + 1);
}
f = ff;
g = gg;
ans = max(ans, max(f, g));
}
return ans;
}
};
Go
func maxNonDecreasingLength(nums1 []int, nums2 []int) int {
n := len(nums1)
f, g, ans := 1, 1, 1
for i := 1; i < n; i++ {
ff, gg := 1, 1
if nums1[i] >= nums1[i-1] {
ff = max(ff, f+1)
}
if nums1[i] >= nums2[i-1] {
ff = max(ff, g+1)
}
if nums2[i] >= nums1[i-1] {
gg = max(gg, f+1)
}
if nums2[i] >= nums2[i-1] {
gg = max(gg, g+1)
}
f, g = ff, gg
ans = max(ans, max(f, g))
}
return ans
}
TypeScript
function maxNonDecreasingLength(nums1: number[], nums2: number[]): number {
const n = nums1.length;
let [f, g, ans] = [1, 1, 1];
for (let i = 1; i < n; ++i) {
let [ff, gg] = [1, 1];
if (nums1[i] >= nums1[i - 1]) {
ff = Math.max(ff, f + 1);
}
if (nums1[i] >= nums2[i - 1]) {
ff = Math.max(ff, g + 1);
}
if (nums2[i] >= nums1[i - 1]) {
gg = Math.max(gg, f + 1);
}
if (nums2[i] >= nums2[i - 1]) {
gg = Math.max(gg, g + 1);
}
f = ff;
g = gg;
ans = Math.max(ans, f, g);
}
return ans;
}