Monotonic Array

Problem Description

An array is monotonic if it is either monotone increasing or monotone decreasing.

An array nums is monotone increasing if for all i <= j, nums[i] <= nums[j]. An array nums is monotone decreasing if for all i <= j, nums[i] >= nums[j].

Given an integer array nums, return true if the given array is monotonic, or false otherwise.

Example

Example 1:

Input: nums = [1,2,2,3]
Output: true

Example 2:

Input: nums = [6,5,4,4]
Output: true

Constraints

• -10^5 <= nums[i] <= 10^5

Solution Approach

Intuition:

To efficiently determine the monotonic array

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

 
class Solution {
isMonotonic(nums) {
 const n = nums.length;
 let increasing = true;
 let decreasing = true;
 for(let i = 0; i < n - 1; i++){
   if(nums[i] > nums[i+1]) increasing = false;
   if(nums[i] < nums[i+1]) decreasing = false;
 }
 return increasing || decreasing;
}
}

TypeScript

Written by @Ishitamukherjee2004

 class Solution {
isMonotonic(nums: number[]): boolean {
 const n = nums.length;
 let increasing = true;
 let decreasing = true;
 for(let i = 0; i < n - 1; i++){
   if(nums[i] > nums[i+1]) increasing = false;
   if(nums[i] < nums[i+1]) decreasing = false;
 }
 return increasing || decreasing;
}
}

Python

Written by @Ishitamukherjee2004

 class Solution:
 def isMonotonic(self, nums):
     n = len(nums)
     increasing = True
     decreasing = True
     for i in range(n - 1):
         if nums[i] > nums[i+1]:
             increasing = False
         if nums[i] < nums[i+1]:
             decreasing = False
     return increasing or decreasing

Java

Written by @Ishitamukherjee2004

 public class Solution {
 public boolean isMonotonic(int[] nums) {
     int n = nums.length;
     boolean increasing = true;
     boolean decreasing = true;
     for(int i = 0; i < n - 1; i++){
         if(nums[i] > nums[i+1]) increasing = false;
         if(nums[i] < nums[i+1]) decreasing = false;
     }
     return increasing || decreasing;
 }
}

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 bool isMonotonic(vector<int>& nums) {
     int n = nums.size();
     bool increasing = true;
     bool decreasing = true;
     
     for(int i=0; i<n-1; i++){
         if(nums[i] > nums[i+1]) increasing = false;
         if(nums[i] < nums[i+1]) decreasing = false;
     }
     
     return increasing || decreasing;
 }
};

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$
• The time complexity is $O(n)$ where n is the length of the input array nums. This is because the algorithm iterates through the array once, performing a constant amount of work for each element.
• The space complexity is $O(1)$ because we are not using any extra space.