Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Rearrange Array Alternately with Positive and Negative Elements
Optimal Approach



Problem Statement

Given an array of integers that contains both positive and negative numbers, your task is to rearrange the elements so that they appear in an alternating order of positive and negative values, starting with a positive number.

If the array is empty or has only one type of number (only positive or only negative), it should be returned as-is.

Examples

Input ArrayRearranged OutputDescription
[1, 2, -3, -4, 5, -6][1, -3, 2, -4, 5, -6]Alternating positive and negative, starting with positive
[-1, 2, -3, 4, -5, 6][2, -1, 4, -3, 6, -5]Signs alternate starting with positive, values rearranged
[1, 2, 3, 4][1, 2, 3, 4]Only positive numbers, no rearrangement needed
[-1, -2, -3, -4][-1, -2, -3, -4]Only negative numbers, no rearrangement needed
[1][1]Single element array, returned as-is
[-1][-1]Single negative number, returned as-is
[][]Empty array, output is also empty
[1, -2, 3, -4, 5, -6, 7][1, -2, 3, -4, 5, -6, 7]One extra positive number at the end
[1, -2, -3, -4][1, -2, -3, -4]More negatives than positives; remaining negatives stay at end

Solution

To solve this problem, we need to rearrange the array such that the signs of the numbers alternate—starting with a positive number, followed by a negative, and so on. The aim is to do this efficiently without using extra space.

We handle this using an index-based placement strategy. We imagine two roles: one index (say posIndex) is in charge of placing positives at even indices (0, 2, 4...), and another index (say negIndex) is for placing negatives at odd indices (1, 3, 5...).

As we go through the array, we check if the current element is at the wrong place based on its sign. If it is, we place it in the correct index (posIndex or negIndex) and increment that index by 2. This ensures alternate placement of positives and negatives.

Handling Different Cases

  • Normal case: When the number of positives and negatives is roughly equal, this alternating pattern works perfectly, and we end up with a well-balanced array.
  • Unequal count: If one type (say, positives) is more than the other, then the extra values will be left over at the end. That’s okay and acceptable by the problem definition.
  • All positive or all negative: If there’s no opposite sign to alternate with, we simply return the array as-is. Nothing can be rearranged here.
  • Empty array: The result is also an empty array since there's nothing to rearrange.
  • Single element: Whether it’s positive or negative, it should be returned untouched because there's no second number to alternate with.

This approach is optimal as it rearranges the array in a single traversal and uses only constant extra space (O(1)).

Also, note that the relative order of the numbers does not matter. So swapping values to place them correctly is allowed and helps us achieve this in-place.

Visualization

Algorithm Steps

  1. Given an array arr of integers containing both positive and negative numbers.
  2. We want to rearrange it such that the signs alternate, starting with a positive number.
  3. Initialize posIndex = 0 and negIndex = 1.
  4. Traverse the array once. For every positive element found at the wrong position (odd index), place it at posIndex and increment posIndex by 2.
  5. For every negative element found at the wrong position (even index), place it at negIndex and increment negIndex by 2.
  6. Repeat the process until both indices exceed the array length.

Code

Python
JavaScript
Java
C++
C
def rearrange_alternating_optimal(arr):
    n = len(arr)
    result = [0] * n
    pos_index, neg_index = 0, 1
    for num in arr:
        if num >= 0 and pos_index < n:
            result[pos_index] = num
            pos_index += 2
        elif num < 0 and neg_index < n:
            result[neg_index] = num
            neg_index += 2
    return result

# Sample Input
arr = [1, 2, 3, -4, -1, 4]
print("Rearranged Array:", rearrange_alternating_optimal(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)In the best case, where the array is already alternating with correct signs at correct indices, we still traverse the array once to validate, resulting in linear time.
Average CaseO(n)Each element in the array is visited once during the traversal to place it in the correct position, ensuring a single-pass solution.
Average CaseO(n)Regardless of the initial arrangement of positive and negative numbers, the array is traversed once and elements are moved in constant time per element.

Space Complexity

O(n)

Explanation: Although the algorithm is optimal in time, it creates a new array to store elements in the correct alternating order, requiring space proportional to the input size.



Welcome to ProgramGuru

Sign up to start your journey with us

Support ProgramGuru.org

You can support this website with a contribution of your choice.

When making a contribution, mention your name, and programguru.org in the message. Your name shall be displayed in the sponsors list.

PayPal

UPI

PhonePe QR

MALLIKARJUNA M