Rearrange Array Alternately with Positive and Negative Elements Optimal Approach

Contents

ProblemExamplesVisualizationSolutionAlgorithmCodeTimeSpace

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 there are extra positive or negative numbers, place them at the end of the array.
  • The relative order of elements is not required to be preserved.

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 Array Rearranged Output Description
[1, 2, -3, -4, 5, -6] [1, -3, 2, -4, 5, -6] Alternating positive and negative, starting with positiveVisualization
[-1, 2, -3, 4, -5, 6] [2, -1, 4, -3, 6, -5] Signs alternate starting with positive, values rearrangedVisualization
[1, 2, 3, 4] [1, 2, 3, 4] Only positive numbers, no rearrangement neededVisualization
[-1, -2, -3, -4] [-1, -2, -3, -4] Only negative numbers, no rearrangement neededVisualization
[1] [1] Single element array, returned as-isVisualization
[-1] [-1] Single negative number, returned as-isVisualization
[] [] Empty array, output is also emptyVisualization
[1, -2, 3, -4, 5, -6, 7] [1, -2, 3, -4, 5, -6, 7] One extra positive number at the endVisualization
[1, -2, -3, -4] [1, -2, -3, -4] More negatives than positives; remaining negatives stay at endVisualization

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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.

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.
Worst 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.