🚨 Uber 2026 OA Full Breakdown: Binary Ops + Monotonic Stack + Balanced Subarray + Digit Swap Pairs — 22 Minutes 4 Problems AC!

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Recently tackled Uber 2026 OA, platform is still CodeSignal, 4 problems, 70 minutes to complete.

Honestly, for those with decent fundamentals, these problems feel very familiar. I finished all of them in just 22 minutes. Two out of four problems were practically free points — you could write the solution at first glance, zero difficulty.

To help everyone practice ahead, I've compiled the approach and solution for each problem.

Uber 2026 OA Interview Overview

Item	Details
Platform	CodeSignal
Questions	4 problems
Time	70 minutes
Difficulty	Medium-high, but some problems can be solved quickly if familiar

Key Focus Areas

Binary operations and bit manipulation
Stack and monotonic stack techniques
Subarray and permutation verification
String operations and swap matching logic

Overall, Uber OA values logical clarity, correct thinking, and implementation speed more than algorithmic complexity, but details are easy to mess up.

Q1: Minimum Operations to Reduce Binary Number to Zero

🔍 Problem Description

Given a positive integer n, each operation can add or subtract 2^i (i ≥ 0). Find the minimum number of operations to reduce n to 0.

📝 Solution Approach

Treat as binary processing
Process each bit from LSB to MSB while maintaining carry
Current bit value = original bit + carry
- If value is 1: can subtract to remove, ops +1, or add to generate carry
- If value is 2 (bit 1 + carry 1): just carry, no operation needed
Accumulate operation count
End when final carry is 0

💡 Key Insight

Consecutive 1s in binary can be optimized via "add 1 to carry" then "subtract"
Example: 111 (7) → add 1 becomes 1000 (8) → subtract 8, total 2 operations
Direct subtraction: 111 → -4 → -2 → -1, needs 3 operations

🧑‍💻 Code Implementation

def min_operations_to_zero(n: int) -> int:
    """
    Uber OA: Minimum operations to reduce binary number to zero
    
    Args:
        n: Positive integer
    
    Returns:
        int: Minimum number of operations
    """
    operations = 0
    carry = 0
    
    while n > 0 or carry > 0:
        # Current bit value = original bit + carry
        current_bit = (n & 1) + carry
        
        if current_bit == 1:
            # Value is 1: need one operation (add or subtract)
            # Greedy: if next bit is also 1, choose add (generate carry, may merge consecutive 1s)
            next_bit = (n >> 1) & 1
            if next_bit == 1:
                # Add: generate carry
                carry = 1
            else:
                # Subtract: no carry
                carry = 0
            operations += 1
        elif current_bit == 2:
            # Value is 2: just carry, no operation
            carry = 1
        else:
            # Value is 0: no operation needed
            carry = 0
        
        n >>= 1
    
    return operations


# Test cases
print(min_operations_to_zero(7))   # 7 = 111 -> +1 = 1000 -> -8 = 0, Answer: 2
print(min_operations_to_zero(5))   # 5 = 101 -> -4 = 1 -> -1 = 0, Answer: 2
print(min_operations_to_zero(1))   # 1 = 1 -> -1 = 0, Answer: 1
print(min_operations_to_zero(15))  # 15 = 1111 -> +1 = 10000 -> -16 = 0, Answer: 2
print(min_operations_to_zero(10))  # 10 = 1010 -> -8 = 2 -> -2 = 0, Answer: 2

⚠️ Tips

The key in binary bit processing is carry maintenance
Must traverse from LSB to MSB, otherwise operation count will be wrong
Drawing binary diagrams helps understand the operation logic quickly

Q2: Discounted Price Sum

🔍 Problem Description

Each item's selling price = original price − the first item to the right with price ≤ current price (sell at original price if none exists).

Find total selling price and indices of items sold at original price.

📝 Solution Approach

Traverse items from right to left
Use monotonic increasing stack to find first item to right with price ≤ current
Stack empty ⇒ sell at original price, record index
Stack not empty ⇒ selling price = original − stack top price
Accumulate total selling price
Output total and original price indices (ascending order)

🧑‍💻 Code Implementation

def discounted_price_sum(prices: list) -> tuple:
    """
    Uber OA: Discounted price sum
    
    Args:
        prices: List of original prices
    
    Returns:
        tuple: (total selling price, list of original price indices)
    """
    n = len(prices)
    total = 0
    full_price_indices = []
    stack = []  # Monotonic increasing stack, stores (price, index)
    
    # Traverse from right to left
    for i in range(n - 1, -1, -1):
        current_price = prices[i]
        
        # Pop all elements greater than current price
        while stack and stack[-1][0] > current_price:
            stack.pop()
        
        if not stack:
            # Stack empty: sell at original price
            total += current_price
            full_price_indices.append(i)
        else:
            # Stack not empty: discount price = original - first ≤ price to right
            discount_price = current_price - stack[-1][0]
            total += discount_price
        
        # Push current element
        stack.append((current_price, i))
    
    # Sort indices ascending
    full_price_indices.sort()
    
    return total, full_price_indices


# Test cases
prices1 = [8, 4, 6, 2, 3]
print(discounted_price_sum(prices1))
# Analysis:
# i=4: 3, stack empty, original 3
# i=3: 2, stack empty, original 2
# i=2: 6, stack top 2 ≤ 6, price 6-2=4
# i=1: 4, stack top 2 ≤ 4, price 4-2=2
# i=0: 8, stack top 4 ≤ 8, price 8-4=4
# Total: 4+2+4+2+3 = 15, original indices: [3, 4]

prices2 = [5, 1, 4, 2]
print(discounted_price_sum(prices2))

⚠️ Tips

Monotonic stack is the classic tool for "first element to right meeting condition"
Core is order and stack state maintenance
Remember to sort indices before output, otherwise may WA

Q3: Balanced Subarray

🔍 Problem Description

For each k (1 to n), determine if there exists a subarray that is exactly a permutation of 1 to k.

📝 Solution Approach

First record each number's position pos[value]
From k = 1, maintain the minimum interval [L, R] containing numbers 1 ~ k
- L = min(pos[1..k])
- R = max(pos[1..k])
If R − L + 1 == k, then k is balanced
- Interval length equals k exactly, and value range is 1 ~ k, must be permutation
Otherwise not balanced

🧑‍💻 Code Implementation

def balanced_subarray(arr: list) -> list:
    """
    Uber OA: Check if 1~k has balanced subarray
    
    Args:
        arr: Permutation of 1~n
    
    Returns:
        list: Boolean array of length n, result[i] indicates if k=i+1 is balanced
    """
    n = len(arr)
    
    # Record each number's position (1-indexed value -> 0-indexed position)
    pos = [0] * (n + 1)
    for i, val in enumerate(arr):
        pos[val] = i
    
    result = []
    min_pos = float('inf')
    max_pos = float('-inf')
    
    for k in range(1, n + 1):
        # Update interval range containing 1~k
        min_pos = min(min_pos, pos[k])
        max_pos = max(max_pos, pos[k])
        
        # Check if interval length exactly equals k
        interval_len = max_pos - min_pos + 1
        if interval_len == k:
            result.append(True)
        else:
            result.append(False)
    
    return result


# Test cases
arr1 = [3, 1, 2, 5, 4]
print(balanced_subarray(arr1))
# k=1: pos[1]=1, interval[1,1], len=1 ✓
# k=2: pos[2]=2, interval[1,2], len=2 ✓
# k=3: pos[3]=0, interval[0,2], len=3 ✓
# k=4: pos[4]=4, interval[0,4], len=5 ≠ 4 ✗
# k=5: pos[5]=3, interval[0,4], len=5 ✓
# Output: [True, True, True, False, True]

arr2 = [2, 1, 4, 3]
print(balanced_subarray(arr2))
# k=1: pos[1]=1, interval[1,1], len=1 ✓
# k=2: pos[2]=0, interval[0,1], len=2 ✓
# k=3: pos[3]=3, interval[0,3], len=4 ≠ 3 ✗
# k=4: pos[4]=2, interval[0,3], len=4 ✓
# Output: [True, True, False, True]

⚠️ Tips

Key is recording number positions and maintaining interval min/max
Can cumulatively update min/max during traversal, no need for repeated scans
This problem is perfect for array + index mapping, time complexity O(n)

Q4: Maximum Pairs by Swapping Digits

🔍 Problem Description

Two numbers can form a pair if they can become each other through at most two swaps of digit positions (including already equal). Find total number of pairs.

📝 Solution Approach

Different lengths ⇒ cannot pair
Same length:
- Convert numbers to strings and sort characters
- If sorted strings are same ⇒ can become each other through swaps (count)
- If sorted strings differ:
  - Count differing positions
  - If diff positions ≤ 4 and chars can match pairwise ⇒ can do two swaps
Implementation:
- Group by length first
- Within group, use hash table to count sorted string frequency for equal pairs
- For unequal cases, check if two-swap condition is met

🧑‍💻 Code Implementation

from collections import defaultdict

def max_pairs_by_swapping(nums: list) -> int:
    """
    Uber OA: Maximum pairs with at most two swaps
    
    Args:
        nums: List of numbers
    
    Returns:
        int: Total number of pairs
    """
    # Group by length
    length_groups = defaultdict(list)
    for num in nums:
        s = str(num)
        length_groups[len(s)].append(s)
    
    total_pairs = 0
    
    for length, group in length_groups.items():
        n = len(group)
        
        # Count strings with same sorted chars (can become each other via swaps)
        sorted_count = defaultdict(int)
        for s in group:
            sorted_s = ''.join(sorted(s))
            sorted_count[sorted_s] += 1
        
        # Calculate pairs with same sorted string: C(count, 2) = count * (count - 1) / 2
        for count in sorted_count.values():
            total_pairs += count * (count - 1) // 2
    
    return total_pairs


def can_match_with_two_swaps(s1: str, s2: str) -> bool:
    """
    Check if two strings can become each other with at most two swaps
    """
    if len(s1) != len(s2):
        return False
    
    # Find differing positions
    diff_positions = []
    for i in range(len(s1)):
        if s1[i] != s2[i]:
            diff_positions.append(i)
    
    # 0 differences: already equal
    if len(diff_positions) == 0:
        return True
    
    # 2 differences: one swap
    if len(diff_positions) == 2:
        i, j = diff_positions
        return s1[i] == s2[j] and s1[j] == s2[i]
    
    # 4 differences: two swaps (need to check if chars match)
    if len(diff_positions) == 4:
        chars1 = [s1[i] for i in diff_positions]
        chars2 = [s2[i] for i in diff_positions]
        return sorted(chars1) == sorted(chars2)
    
    return False


# More precise version (checks each pair for swap possibility)
def max_pairs_precise(nums: list) -> int:
    """
    Precise version: check if each pair can match with at most two swaps
    """
    n = len(nums)
    pairs = 0
    
    for i in range(n):
        for j in range(i + 1, n):
            s1, s2 = str(nums[i]), str(nums[j])
            if can_match_with_two_swaps(s1, s2):
                pairs += 1
    
    return pairs


# Test cases
nums1 = [123, 321, 132, 456]
print(max_pairs_by_swapping(nums1))  # 123, 321, 132 can pair with each other, 3 pairs

nums2 = [111, 111, 111]
print(max_pairs_by_swapping(nums2))  # 3 identical, C(3,2) = 3 pairs

nums3 = [12, 21, 34, 43, 56]
print(max_pairs_by_swapping(nums3))  # (12,21), (34,43) = 2 pairs

⚠️ Tips

Sorting string chars quickly determines if swap transformation is possible
Diff positions ≤ 4 is the key condition
Hash table grouping significantly reduces time complexity

Complexity Analysis

Problem	Time Complexity	Space Complexity
Q1: Binary to Zero	O(log n)	O(1)
Q2: Discounted Price	O(n)	O(n)
Q3: Balanced Subarray	O(n)	O(n)
Q4: Max Pairs	O(n² × L) or O(n × L log L)	O(n × L)

Where L = maximum number of digits

🔥 Why Most People Fail

Common Mistake	Correct Approach
Q1 forget carry handling	Maintain carry variable
Q1 process from MSB	Must traverse from LSB to MSB
Q2 wrong monotonic stack direction	Traverse from right to left
Q2 indices not sorted	Sort ascending before output
Q3 repeated scanning for min/max	Cumulatively update min_pos, max_pos
Q4 only consider equal case	Also consider 2 and 4 diff positions

FAQ | Uber 2026 OA Common Questions

Q1: How many problems in Uber 2026 OA? What's the difficulty?

Uber 2026 OA has 4 problems, including binary min operations, monotonic stack discount price, subarray permutation check, and max digit swap pairs. Difficulty is medium-high, but most are common template problems. With familiarity with CodeSignal and Python solving approaches, you can usually finish quickly.

Q2: How long does Uber 2026 OA take to complete?

Based on real experience, if familiar with problem types, all four can be done in 20-30 minutes. I personally finished the whole set in 22 minutes first try. Proper problem ordering and using Python template code can significantly save time.

Q3: What are Uber 2026 OA's focus areas?

Main abilities tested:

Binary bit operations and carry handling
Monotonic stack application and discount calculation logic
Subarray permutation check and interval min/max maintenance
String-number matching and at-most-two-swap logic

Also tests edge case handling, complexity understanding, and implementation ability.

Q4: What are Python's advantages in Uber OA?

Python's concise syntax and built-in data structures (dict, list, set) and functions (min, max, sorted) are perfect for OA logic, especially:

Binary bit operations and carry simulation
Monotonic stack operations
Index mapping and subarray checking
String conversion and hash counting

Using Python templates properly improves speed and accuracy.

Q5: Any tips for passing Uber OA quickly?

Summary:

Familiarize with template types: binary processing, monotonic stack, interval checking, char swap matching
Clarify logic before coding: sketch or test with small examples
Allocate time wisely: do familiar problems first, leave hard ones for last
Review edge cases: especially swap count, carry, index bounds, char matching
Reuse template code: prepare utility functions in advance

Q6: Do Uber OA problems repeat?

Based on real experience, CodeSignal platform problems have high type repetition, especially classic templates like binary operations, monotonic stack, and subarray permutations. Practicing these types ahead and understanding general solutions and edge handling greatly improves AC rate.

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