# Check If The Given Array is Monotonic

An array is called monotonic if the index of each element increases from the first to the last or decreases from the first to the last.

## Table of Contents

## Introduction

A monotonic array is an array it is either increasing or monotonic decreasing.

## Problem Statement

An array `A`

is monotone increasing(📈) if for all `i <= j`

, `A[i] <= A[j]`

, where `i`

and `j`

are index elements.

An array `A`

is monotone decreasing (📉) if for all `i >= j`

, `A[i] >= A[j]`

. `true`

is returned if and only if the given array `A`

is monotonic.

## Explanation:

__Monotone increasing__: An array `A`

is called monotone increasing if the first number is less than or equal to the second number, the second number is less than or equal to the third number, and so on, and vice versa.

__Monotone decreasing__: An array `A`

is called monotone decreasing if the first number is greater than or equal to the second number, the second number is greater than or equal to the third number, and so on, and vice versa.

**Example 1:**

```
Input: [1, 2, 2, 3]
Output: true
```

**Example 2:**

```
Input: [1, 2, 6, 2, 3]
Output: false
```

**Example 3:**

```
Input: [7, 2, 1]
Output: true
```

## Thought Process

An array is called monotonic if the index of each element increases from the first to the last or decreases from the first to the last.

## Algorithm

We need to run two for loops to check if either of the loops returns true.

We must check if the previous index is less than the current one for the monotonic increasing array. And for the monotonic decreasing array, we must check if the previous index is greater than the current one.

Finally, we return `true`

if either of the loops evaluates to `true`

.

: we can do using one pass but let us look at both the algorithms.Optimal way

## Intuition

There are different ways to solve this problem, but the one that is simple yet powerful are:

- Two-pass approach
- One-pass approach

## Two-pass approach

To check if the elements are in increasing or decreasing order, you need to run two separate loops.

### Code

```
class MonotonicArray {
public static void main(String[] args) {
int[] input = {1, 2, 2, 3};
System.out.println(isMonotonic(input)); // true
}
public static boolean isMonotonic(int[] array) {
return isIncreasing(array) || isDecreasing(array);
}
public static boolean isIncreasing(int[] nums) {
for (int i = 1; i < nums.length; i++)
if (nums[i - 1] > nums[i]) {
return false;
}
return true;
}
public static boolean isDecreasing(int[] nums) {
for (int i = 1; i < nums.length; i++)
if (nums[i - 1] < nums[i]) {
return false;
}
return true;
}
}
```

### Complexity Analysis

__Time complexity__: We have two for-loops running`O(n)`

times each. So the overall time complexity is`O(n) + O(n) = 2*O(n)`

. Hence the time is`O(n)`

for this algorithm.__Space complexity__:`O(1)`

, no extra space is used for this algorithm.

Let us optimize the above code snippet with a single loop.

## One-Pass approach

The above algorithm can be optimized to a single loop using `boolean`

flags.

### Code

```
class MonotonicArray {
public static void main(String[] args) {
int[] input = {1, 2, 2, 3};
System.out.println(isMonotonic(input)); // true
}
public static boolean isMonotonic(int[] array) {
boolean isIncreasing = true;
boolean isDecreasing = true;
for (int i = 1; i < array.length; i++) {
if (array[i] < array[i - 1]) {
isDecreasing = false;
}
if (array[i] > array[i - 1]) {
isIncreasing = false;
}
}
return isIncreasing || isDecreasing;
}
}
```

### Complexity Analysis

__Time complexity__: We have one for-loop running`O(n)`

time. So, the overall time complexity is`O(n)`

for this algorithm.__Space complexity__:`O(1)`

, no extra space is used for this algorithm.

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