Given an array of integers, write a function to optimize the inversion count without using recursion. An inversion is defined as a pair of indices (i, j) such that i arr[j]. Your function should return the total number of inversions in the array, with a time complexity better than O(n^2).

In this problem, we need to count the number of inversions in an array efficiently. An inversion is defined as a pair of indices (i, j) such that i arr[j]. The brute-force approach would involve checking all pairs of indices, which results in a time complexity of O(n^2). However, we can utilize the 'Merge Sort' algorithm to achieve a time complexity of O(n log n). This is because during the merge step of Merge Sort, we can count how many elements from the right half of the array are less than elements from the left half, which directly translates to counting inversions. We will modify the Merge Sort algorithm to count inversions as follows: 1. **Divide**: Split the array into two halves. 2. **Conquer**: Recursively sort both halves while counting inversions in each half. 3. **Combine**: During the merge step, count how many elements from the right half are placed before elements from the left half (this indicates inversions). For example, if the left half has elements [2, 4] and the right half has [1, 3], when merging, '1' is less than both '2' and '4', yielding two inversions. 4. Return the total count of inversions combined from both halves and during the merge. Here is the Python implementation of the above approach: ```python def merge_and_count(arr, temp_arr, left, mid, right): i = left # Starting index for left subarray j = mid + 1 # Starting index for right subarray k = left # Starting index to be sorted inv_count = 0 while i <= mid and j <= right: if arr[i] <= arr[j]: temp_arr[k] = arr[i] i += 1 else: temp_arr[k] = arr[j] inv_count += (mid - i + 1) # Count inversions j += 1 k += 1 while i <= mid: temp_arr[k] = arr[i] i += 1 k += 1 while j <= right: temp_arr[k] = arr[j] j += 1 k += 1 for i in range(left, right + 1): arr[i] = temp_arr[i] return inv_count def merge_sort_and_count(arr, temp_arr, left, right): inv_count = 0 if left < right: mid = (left + right) // 2 inv_count += merge_sort_and_count(arr, temp_arr, left, mid) inv_count += merge_sort_and_count(arr, temp_arr, mid + 1, right) inv_count += merge_and_count(arr, temp_arr, left, mid, right) return inv_count def count_inversions(arr): temp_arr = [0] * len(arr) return merge_sort_and_count(arr, temp_arr, 0, len(arr) - 1) ``` This code first defines a helper function for merging and counting inversions, then uses the modified Merge Sort to count inversions in the entire array. The time complexity of this solution is O(n log n) due to the Divide and Conquer approach of Merge Sort. Each merge operation takes linear time, and the levels of recursion are logarithmic based on the size of the array. The space complexity is O(n) due to the temporary array used for merging. Given the constraints of the problem, this solution is efficient and optimal. 1. **Empty Array**: If the input array is empty, the function should return 0 since there are no elements to compare. 2. **Single Element Array**: An array with a single element should also return 0 as there are no pairs. 3. **All Elements Same**: An array where all elements are the same (e.g., [1, 1, 1]) should return 0 as no inversions exist. 4. **Already Sorted Array**: An already sorted array (e.g., [1, 2, 3, 4]) should return 0. 5. **Worst Case (Descending Order)**: An array sorted in descending order (e.g., [5, 4, 3, 2, 1]) should return the maximum number of inversions, which is n(n-1)/2, where n is the number of elements.

Optimize inversion for without recursion | Postmates Software Engineer Interview

Optimize inversion for without recursion

Last updated: November 6, 2025

Quick Overview

Given an array of integers, write a function to optimize the inversion count without using recursion. An inversion is defined as a pair of indices (i, j) such that i < j and arr[i] > arr[j]. Your function should return the total number of inversions in the array, with a time complexity better than O(n^2).

Postmates

Coding & Algorithms

Software Engineer

Postmates

November 6, 2025

Software Engineer

Onsite

Coding & Algorithms

Medium

3,399 solved

Coding interviews at Postmates focus on problem-solving approach as much as the final solution. The interviewer wants to see you break down the problem, consider edge cases, and optimize iteratively. Communication throughout the process is key.

What the Interviewer Expects

Recognize the underlying problem pattern (sliding window, two pointers, BFS/DFS, etc.)
Discuss multiple approaches and trade-offs before coding
Implement an optimal solution with clean, production-quality code
Handle all edge cases including boundary conditions and invalid input
Optimize both time and space complexity with clear justification
Test your solution systematically with well-chosen examples

Key Topics to Cover

Edge cases and input validation

Graph algorithms and traversal

Hash maps and frequency counting

Common algorithm patterns (sliding window, two pointers, BFS/DFS)

Tree structures and recursion

How to Approach This

Clarify input constraints and edge cases before writing code.
Walk through your approach verbally and confirm with the interviewer before coding.
Start with a brute force solution, then optimize. Mention time and space complexity.
Test your solution with examples, including edge cases like empty input or duplicates.
Consider common patterns: sliding window, two pointers, hash map, BFS/DFS, dynamic programming.

Possible Follow-up Questions

What happens if the input contains duplicates?
What if the input doesn't fit in memory?
Can you solve this in a single pass?

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Problem Analysis

In this problem, we need to count the number of inversions in an array efficiently. An inversion is defined as a pair of indices (i, j) such that i < j and arr[i] > arr[j]. The brute-force approach wo...

Approach

We will modify the Merge Sort algorithm to count inversions as follows:

Divide: Split the array into two halves.
Conquer: Recursively sort both halves while counting inversions in each h...

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Optimize inversion for without recursion

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