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
.
Optimal way: we can do using one pass but let us look at both the algorithms.
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 isO(n) + O(n) = 2*O(n)
. Hence the time isO(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 isO(n)
for this algorithm. - Space complexity:
O(1)
, no extra space is used for this algorithm.
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