Difficulty: Easy, Asked-in: Google, Amazon, Adobe
Key takeaway: An excellent problem to learn problem-solving using sorting and single scan (Incrementing the loop by 2). Such problems are popular where we need to rearrange the input in place to get the output.
Given an unsorted array of n integers, write a program to sort the array into a wave array. An array A[n] is sorted in wave arrangement if A[0] >= A[1] <= A[2] >= A[3] <= A[4] >= …. Note: There can be multiple possible answers, but we need to return any one of the possible waveforms.
Example 1
Input: A[] = [1, 2, 3, 4]
Output: A[] = [2, 1, 4, 3] or [4, 1, 3, 2] or any other wave form like structure.
Example 2
Input: A[] = [20, 10, 8, 6, 4, 2]
Output: A[] = [20, 8, 10, 4, 6, 2] or [10, 8, 20, 2, 6, 4] or any other wave form like structure.
If we look at the wave-like output, values are organized in an alternate order of large and small values. So how do we transform the given input in such an order? One idea is: If we sort the input array, can we arrange the elements in waveform? Let’s think!
In sorted array, all elements will be arranged in increasing order, i.e, A[0] ≤ A[1] ≤ A[2] ≤ A[3]≤ ….≤A[n - 2] ≤ A[n - 1]. So if we pick elements in pairs and swap the adjacent elements, the sorted array will get arranged in wave order. Think!
Visualization of sorted array
Visualization after swapping adjacent elements in the sorted array
// Function to convert the given array to wave form
void waveArray(int A[], int n)
{
// sort the array
sort(A, A + n);
// swap adjacent elements
for (int i = 0; i < n - 1; i = i + 2)
swap(A[i], A[i + 1]);
}
def waveArray(A):
# sort the array
A.sort()
# swap adjacent elements
for i in range(0, len(A) - 1, 2):
A[i], A[i + 1] = A[i + 1], A[i]
Suppose we use an O(nlogn) sorting algorithm like quicksort, merge sort, or heap sort. Time complexity = Time complexity of sorting + Time complexity of swapping adjacent elements = O(nlogn) + O(n) = O(nlogn)
Space complexity is O(1) if we use heap sort and O(n) if we use merge sort.
The critical question is: Can we solve this problem in O(n) time complexity, i.e., without sorting? Let's think and explore one level further!
From the given output pattern, if we ensure that values at all the even positions are greater than the odd position, we can achieve the wave-like structure. In other words, for every even index i, we need to ensure this order in triplet (A[i - 1], A[i], A[i + 1]) => A[i - 1] < A[i] > A[i + 1].
There can be four possible cases for each triplet in the input array:
If we observe, we can simplify further from cases 1, 2, and 3. The idea is simple: we need to place the max value of each triplet in the middle.
To implement this, we can run a loop incrementing by 2 to compare adjacent elements and ensure that all elements at even positions are greater than their adjacent elements at odd positions.
If current element A[i] is smaller than previous element A[i - 1], we swap current element A[i] with A[i - 1]. Note: We should avoid this operation for i = 0 because there is no previous element.
if(i > 0 && A[i - 1] > A[i])
swap(A[i], A[i - 1])
If current element A[i] is smaller than next element A[i + 1], we swap current element A[i] with A[i + 1]. Note: We should avoid this operation for i = n - 1 because there is no next element.
if(i < n - 1 && A[i] < A[i + 1])
swap(A[i], A[i + 1])
void waveArray(int A[], int n)
{
for(int i = 0; i < n; i = i + 2)
{
if(i > 0 && A[i - 1] > A[i])
swap(A[i], A[i - 1])
if(i < n - 1 && A[i] < A[i + 1])
swap(A[i], A[i + 1])
}
}
// Function to convert the given array to wave form
void waveArray(int A[], int n)
{
// swap elements at even positions
// with the element immediately following it
// if it is greater
for (int i = 0; i < n; i = 1 + 2)
{
if (i > 0 && A[i - 1] > A[i])
swap(A[i], A[i - 1]);
if (i < n - 1 && A[i] < A[i + 1])
swap(A[i], A[i + 1]);
}
}
def waveArray(A):
# swap elements at even positions
# with the element immediately following it
# if it is greater
for i in range(0, len(A), 2):
if i > 0 and A[i - 1] > A[i]:
A[i], A[i - 1] = A[i - 1], A[i]
if i < len(A) - 1 and A[i] < A[i + 1]:
A[i], A[i + 1] = A[i + 1], A[i]
The time complexity for the above solution would be O(n) because we are linearly traversing the array and performing O(1) operations at each iteration. Space complexity = O(1) because we are using constant extra memory.
Thanks to Navtosh Kumar for his contribution in creating the first version of the content. Please write in the message below if you find an error or want to share more insights. Enjoy learning, Enjoy coding, Enjoy algorithms!
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