Even-Odd Extremes Difference — Problem Statement & Solution Guide
Problem Description
Given an integer array nums of size at least 2, find the absolute difference between the maximum value located at an even index and the minimum value located at an odd index.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Even-Odd Extremes Difference"
WHY DOES IT MATTER?
This pattern reduces memory overhead by avoiding auxiliary data structures, which is crucial when dealing with large datasets or streaming data. It also simplifies the code, making it easier to reason about correctness and performance.
OPTIMIZATION CHALLENGE
The insight is that extrema can be updated incrementally without storing all elements, turning a potentially O(n) space solution into an O(1) space one.
REAL-WORLD CONNECTION
In distributed log aggregation, you often need to find the maximum latency per server and the minimum latency per client in a single pass over a massive log stream, mirroring this single-pass extrema pattern.
When explaining this to an interviewer, emphasize the importance of state variables and how they enable constant space, and be ready to discuss edge cases like all even or all odd indices.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the absolute difference between two specific extrema: the maximum value among elements at even indices and the minimum value among elements at odd indices. A naive solution would involve two nested loops or separate scans that collect all even-indexed and odd-indexed values into auxiliary arrays, then compute the extrema, leading to O(n) time but also O(n) additional space. However, the optimal approach requires only a single pass through the array, maintaining two running variables—one for the current maximum at even indices and one for the current minimum at odd indices—thereby achieving O(n) time with O(1) extra space.
The key algorithmic insight is that the extrema can be updated incrementally: when encountering an element at an even index, compare it with the current even-max and update if larger; similarly, when encountering an odd index, compare with the current odd-min and update if smaller. This eliminates the need to store all values, which is especially important for large inputs where memory usage becomes a bottleneck. The final absolute difference is then computed in constant time after the single traversal.
This pattern exemplifies the classic “single-pass scan with running aggregates” paradigm, which is ubiquitous in interview questions involving extrema, prefix sums, or sliding windows. It demonstrates how careful state management can reduce both time and space complexity, a skill that is highly valued in production systems where performance and resource constraints are critical.
Interview Questions on This Problem
Q1How would you modify the algorithm if the array could contain negative numbers and you needed the difference between the largest even-indexed value and the smallest odd-indexed value?
The algorithm remains unchanged; the comparison operators handle negative values naturally. You still track the maximum at even indices and the minimum at odd indices, then compute the absolute difference.
Q2A fintech platform asks: if the array represents transaction amounts over days, how could you extend this solution to also return the indices of the max even and min odd values?
Maintain two additional variables to store the indices alongside the extrema values. Update them whenever you update the corresponding value during the single pass.
Q3During a high-growth startup interview, you’re asked: what if the array is extremely large and streamed in chunks? How would you adapt the solution?
Process each chunk sequentially, updating the running maximum and minimum as you go. Since you only need the final extrema, you can discard processed chunks, keeping the space usage constant.
Examples
Input
[5, 2, 9, 1, 7, 3]
Output
8
Explanation: Step-by-step: with input [5, 2, 9, 1, 7, 3], we first identify the maximum value at an even index, which is 9 (at index 2), and the minimum value at an odd index, which is 1 (at index 3). The absolute difference between these two values is |9 - 1| = 8.
Input
[1, -1, 3, 5, 7, 9]
Output
10
Explanation: Step-by-step: with input [1, -1, 3, 5, 7, 9], we first identify the maximum value at an even index, which is 9 (at index 5), and the minimum value at an odd index, which is -1 (at index 1). The absolute difference between these two values is |9 - (-1)| = 10.
Constraints
- 2 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Traverse the array once, updating two running variables for the even-index maximum and odd-index minimum. After the traversal, compute the absolute difference of these two values.
Brute Force Approach
Collect all even-indexed elements into one list and all odd-indexed elements into another, then compute the max of the first list and the min of the second list, finally take their absolute difference. This requires two passes and extra memory.
Verified Code Solutions
function solution(nums) { let maxEven = -Infinity, minOdd = Infinity; for (let i = 0; i < nums.length; i++) { if (i % 2 === 0 && nums[i] > maxEven) maxEven = nums[i]; if (i % 2 !== 0 && nums[i] < minOdd) minOdd = nums[i]; } return Math.abs(maxEven - minOdd); }class Solution { public: int solution(vector<int>& nums) { int maxEven = INT_MIN, minOdd = INT_MAX; for (int i = 0; i < nums.size(); i++) { if (i % 2 == 0 && nums[i] > maxEven) maxEven = nums[i]; if (i % 2 != 0 && nums[i] < minOdd) minOdd = nums[i]; } return abs(maxEven - minOdd); } };class Solution { public int solution(int[] nums) { int maxEven = Integer.MIN_VALUE, minOdd = Integer.MAX_VALUE; for (int i = 0; i < nums.length; i++) { if (i % 2 == 0 && nums[i] > maxEven) maxEven = nums[i]; if (i % 2 != 0 && nums[i] < minOdd) minOdd = nums[i]; } return Math.abs(maxEven - minOdd); } }def solution(nums): max_even = float('-inf'); min_odd = float('inf'); for i, num in enumerate(nums): if i % 2 == 0 and num > max_even: max_even = num; if i % 2 != 0 and num < min_odd: min_odd = num; return abs(max_even - min_odd)function solution(nums) { let maxEven = -Infinity, minOdd = Infinity; for (let i = 0; i < nums.length; i++) { if (i % 2 === 0 && nums[i] > maxEven) maxEven = nums[i]; if (i % 2 !== 0 && nums[i] < minOdd) minOdd = nums[i]; } return Math.abs(maxEven - minOdd); }Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.