Check if an array is Wave Array
Last Updated :
02 May, 2025
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Given an array of N positive integers. The task is to check if the array is sorted in wave form.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: NOInput: arr[] = {1, 5, 3, 7, 2, 8, 6}
Output: YES
Approach:
- First check the element at index 1, i.e, arr[1] and observe the pattern.
- If arr[1] is greater than its left and right element, then this pattern will be followed by other elements.
- Else If arr[1] is smaller than its left and right element, then this pattern will be followed by other elements.
- Check for the same pattern found from above steps. If at any point, this rule violates, return false, else return true.
Below is the implementation of above approach:
// CPP code to check if the array is wave array
#include <iostream>
using namespace std;
// Function to check if array is wave array
// arr : input array
// n : size of array
bool isWaveArray(int arr[], int n)
{
bool result = false;
/* Check the wave form
* If arr[1] is greater than left and right
* Same pattern will be followed by whole
* elements, else reverse pattern
* will be followed by array elements
*/
if (arr[1] > arr[0] && arr[1] > arr[2]) {
for (int i = 1; i < n - 1; i += 2) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] <= arr[n - 2]) {
result = false;
}
}
}
else if (arr[1] < arr[0] && arr[1] < arr[2]) {
for (int i = 1; i < n - 1; i += 2) {
if (arr[i] < arr[i - 1] && arr[i] < arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] >= arr[n - 2]) {
result = false;
}
}
}
return result;
}
// Driver Code
int main()
{
// Array
int arr[] = { 1, 3, 2, 4 };
int n = sizeof(arr) / sizeof(int);
if (isWaveArray(arr, n)) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
return 0;
}
// Java code to check if the array is wave array
public class GFG {
// Function to check if array is wave array
// arr : input array
// n : size of array
static boolean isWaveArray(int arr[], int n)
{
boolean result = true;
/* Check the wave form
* If arr[1] is greater than left and right
* Same pattern will be followed by whole
* elements, else reverse pattern
* will be followed by array elements
*/
if (arr[1] > arr[0] && arr[1] > arr[2]) {
for (int i = 1; i < n - 1; i += 2) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] <= arr[n - 2]) {
result = false;
}
}
}
else if (arr[1] < arr[0] && arr[1] < arr[2]) {
for (int i = 1; i < n - 1; i += 2) {
if (arr[i] < arr[i - 1] && arr[i] < arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] >= arr[n - 2]) {
result = false;
}
}
}
return result;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 3, 2, 4 };
int n = arr.length;
if (isWaveArray(arr, n)) {
System.out.println("YES");
}
else {
System.out.println("NO");
}
}
// This Code is contributed by ANKITRAI1
}
# Python 3 code to check if
# the array is wave array
# Function to check if
# array is wave array
# arr : input array
# n : size of array
def isWaveArray(arr , n):
result = True
# Check the wave form
# If arr[1] is greater than
# left and right. Same pattern
# will be followed by whole
# elements, else reverse pattern
# will be followed by array elements
if (arr[1] > arr[0] and arr[1] > arr[2]):
for i in range(1, n - 1, 2):
if (arr[i] > arr[i - 1] and
arr[i] > arr[i + 1]):
result = True
else :
result = False
break
# Check for last element
if (result == True and n % 2 == 0):
if (arr[n - 1] <= arr[n - 2]) :
result = False
elif (arr[1] < arr[0] and
arr[1] < arr[2]) :
for i in range(1, n - 1, 2) :
if (arr[i] < arr[i - 1] and
arr[i] < arr[i + 1]):
result = True
else :
result = False
break
# Check for last element
if (result == True and n % 2 == 0) :
if (arr[n - 1] >= arr[n - 2]) :
result = False
return result
# Driver Code
if __name__ == "__main__":
# Array
arr = [ 1, 3, 2, 4 ]
n = len(arr)
if (isWaveArray(arr, n)):
print("YES")
else:
print("NO")
# This code is contributed
# by ChitraNayal
// C# code to check if the
// array is wave array
using System;
class GFG
{
// Function to check if array
// is wave array
// arr : input array
// n : size of array
static bool isWaveArray(int []arr, int n)
{
bool result = true;
/* Check the wave form
* If arr[1] is greater than left
* and right. Same pattern will be
* followed by whole elements, else
* reverse pattern will be followed
by array elements */
if (arr[1] > arr[0] && arr[1] > arr[2])
{
for (int i = 1; i < n - 1; i += 2)
{
if (arr[i] > arr[i - 1] &&
arr[i] > arr[i + 1])
{
result = true;
}
else
{
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0)
{
if (arr[n - 1] <= arr[n - 2])
{
result = false;
}
}
}
else if (arr[1] < arr[0] &&
arr[1] < arr[2])
{
for (int i = 1; i < n - 1; i += 2)
{
if (arr[i] < arr[i - 1] &&
arr[i] < arr[i + 1])
{
result = true;
}
else
{
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0)
{
if (arr[n - 1] >= arr[n - 2])
{
result = false;
}
}
}
return result;
}
// Driver code
public static void Main()
{
int []arr = { 1, 3, 2, 4 };
int n = arr.Length;
if (isWaveArray(arr, n))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
// This code is contributed
// by inder_verma
<script>
// Javascript code to check if the array is wave array
// Function to check if array is wave array
// arr : input array
// n : size of array
function isWaveArray(arr, n)
{
let result = true;
/* Check the wave form
* If arr[1] is greater than left and right
* Same pattern will be followed by whole
* elements, else reverse pattern
* will be followed by array elements
*/
if (arr[1] > arr[0] && arr[1] > arr[2]) {
for (let i = 1; i < n - 1; i += 2) {
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] <= arr[n - 2]) {
result = false;
}
}
}
else if (arr[1] < arr[0] && arr[1] < arr[2]) {
for (let i = 1; i < n - 1; i += 2) {
if (arr[i] < arr[i - 1] && arr[i] < arr[i + 1]) {
result = true;
}
else {
result = false;
break;
}
}
// Check for last element
if (result == true && n % 2 == 0) {
if (arr[n - 1] >= arr[n - 2]) {
result = false;
}
}
}
return result;
}
// Array
let arr = [ 1, 3, 2, 4 ];
let n = arr.length;
if (isWaveArray(arr, n)) {
document.write("YES");
}
else {
document.write("NO");
}
// This code is contributed by divyeshrabadiya07.
</script>
<?php
// PHP code to check if the array is wave array
// Function to check if array is wave array
// arr : input array
// n : size of array
function isWaveArray( $arr, $n)
{
$result = true;
/* Check the wave form
* If arr[1] is greater than left and right
* Same pattern will be followed by whole
* elements, else reverse pattern
* will be followed by array elements
*/
if ($arr[1] > $arr[0] &&
$arr[1] > $arr[2])
{
for ( $i = 1; $i < ($n - 1); $i += 2)
{
if ($arr[$i] > $arr[$i - 1] &&
$arr[$i] > $arr[$i + 1])
{
$result = true;
}
else
{
$result = false;
break;
}
}
// Check for last element
if ($result == true && $n % 2 == 0)
{
if ($arr[$n - 1] <= $arr[$n - 2])
{
$result = false;
}
}
}
else if ($arr[1] < $arr[0] &&
$arr[1] < $arr[2])
{
for ($i = 1; $i < $n - 1; $i += 2)
{
if ($arr[$i] < $arr[$i - 1] &&
$arr[$i] < $arr[$i + 1])
{
$result = true;
}
else
{
$result = false;
break;
}
}
// Check for last element
if ($result == true && $n % 2 == 0)
{
if ($arr[$n - 1] >= $arr[$n - 2])
{
$result = false;
}
}
}
return $result;
}
// Driver Code
// Array
$arr = array (1, 3, 2, 4 );
$n = sizeof($arr);
if (isWaveArray($arr, $n))
{
echo "YES";
}
else
{
echo "NO";
}
// This code is contributed by jit_t
?>
Output
YES
Time Complexity: O(n)
Auxiliary Space: O(1)
Approach :- 2 One approach to check if an array is a wave array is to first sort the array in ascending order. Then, we can swap adjacent elements to form a wave-like pattern. If at least one element does not satisfy the wave property, the array is not a wave array.
Here is the implementation of this approach:
Traverse the array from the second element to the second last element.
Check if the current element is greater than or equal to both its adjacent elements or smaller than or equal to both its adjacent elements.
If it satisfies the above condition, move to the next element, else return false.
If the entire array has been traversed without any element failing the condition, return true.
#include <bits/stdc++.h>
using namespace std;
bool isWave(int arr[], int n) {
// sort the array
sort(arr, arr + n);
// swap adjacent elements to form wave
for (int i = 0; i < n - 1; i += 2) {
swap(arr[i], arr[i + 1]);
}
// check if wave property is satisfied
for (int i = 0; i < n - 1; i++) {
if ((i % 2 == 0 && arr[i] > arr[i + 1]) ||
(i % 2 == 1 && arr[i] < arr[i + 1])) {
return true;
}
}
return false;
}
int main() {
int arr[] = { 1, 3, 2, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
if (isWave(arr, n)) {
cout << "The array is a wave array";
} else {
cout << "The array is not a wave array";
}
return 0;
}
import java.util.Arrays;
public class GFG {
// Function to check if the given array is a wave array
static boolean isWave(int[] arr, int n) {
// Sort the array in ascending order
Arrays.sort(arr);
// Swap adjacent elements to form a wave
for (int i = 0; i < n - 1; i += 2) {
swap(arr, i, i + 1);
}
// Check if the wave property is satisfied
for (int i = 0; i < n - 1; i++) {
// For even indices, check if the current element is greater than the next element
// For odd indices, check if the current element is smaller than the next element
if ((i % 2 == 0 && arr[i] > arr[i + 1]) || (i % 2 == 1 && arr[i] < arr[i + 1])) {
return true; // If the wave property is satisfied, return true
}
}
return false; // If the wave property is not satisfied for any pair, return false
}
// Helper function to swap two elements in an array
static void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public static void main(String[] args) {
int[] arr = { 1, 3, 2, 4 };
int n = arr.length;
if (isWave(arr, n)) {
System.out.println("The array is a wave array");
} else {
System.out.println("The array is not a wave array");
}
}
}
def is_wave(arr):
# Sort the array in ascending order
arr.sort()
# Swap adjacent elements to form a wave
for i in range(0, len(arr) - 1, 2):
arr[i], arr[i + 1] = arr[i + 1], arr[i]
# Check if the wave property is satisfied
for i in range(len(arr) - 1):
if (i % 2 == 0 and arr[i] > arr[i + 1]) or (i % 2 == 1 and arr[i] < arr[i + 1]):
return True
return False
# Driver code
if __name__ == "__main__":
arr = [1, 3, 2, 4]
if is_wave(arr):
print("The array is a wave array")
else:
print("The array is not a wave array")
using System;
public class GFG
{
// Function to check if the array is a wave array
public static bool IsWave(int[] arr, int n)
{
// Sort the array
Array.Sort(arr);
// Swap adjacent elements to form a wave
for (int i = 0; i < n - 1; i += 2)
{
int temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
// Check if the wave property is satisfied
for (int i = 0; i < n - 1; i++)
{
if ((i % 2 == 0 && arr[i] > arr[i + 1]) ||
(i % 2 == 1 && arr[i] < arr[i + 1]))
{
return true;
}
}
return false;
}
public static void Main(string[] args)
{
int[] arr = { 1, 3, 2, 4 };
int n = arr.Length;
if (IsWave(arr, n))
{
Console.WriteLine("The array is a wave array");
}
else
{
Console.WriteLine("The array is not a wave array");
}
}
}
function isWave(arr) {
// Sort the array
arr.sort((a, b) => a - b);
// Swap adjacent elements to form a wave
for (let i = 0; i < arr.length - 1; i += 2) {
[arr[i], arr[i + 1]] = [arr[i + 1], arr[i]];
}
// Check if the wave property is satisfied
for (let i = 0; i < arr.length - 1; i++) {
if ((i % 2 === 0 && arr[i] > arr[i + 1]) || (i % 2 === 1 && arr[i] < arr[i + 1])) {
return true;
}
}
return false;
}
//Driver code
const arr = [1, 3, 2, 4];
if (isWave(arr)) {
console.log("The array is a wave array");
} else {
console.log("The array is not a wave array");
}
Output
The array is a wave array
Time Complexity:
The algorithm involves a linear scan of the entire array, which takes O(nlogn) time, where n is the size of the array.
Auxiliary Space:
The algorithm uses only constant extra space to store the indices and variables, which does not depend on the size of the array. Hence, the space complexity is O(1).
