Dynamic Interval Partition — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing a sequence of data points. Your task is to compute the total sum of all elements in the array. However, the summation must be performed using a specific pairing strategy: start by summing the first and last elements, then the second and second-to-last elements, and so on, moving inward from both ends. If the array contains an odd number of elements, the middle element is added to the total sum as a standalone term. Return the final computed sum.
This problem requires careful index management to ensure that each pair is correctly identified and summed without double-counting or missing elements. The pairing strategy ensures that the summation is performed in a symmetric manner, which can be useful in certain optimization or partitioning scenarios.
The solution must be efficient, with a time complexity of O(n), where n is the length of the array, and a space complexity of O(1), as no additional data structures are required beyond the input array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Interval Partition"
WHY DOES IT MATTER?
The two‑pointer pattern transforms problems that appear to need nested iteration into linear scans, dramatically cutting runtime. It is especially valuable when the input size is large and memory is constrained, as it avoids auxiliary containers.
OPTIMIZATION CHALLENGE
The insight that each index i has a unique counterpart n‑1‑i eliminates redundant work; by moving both pointers inward simultaneously, we guarantee each element is processed exactly once.
REAL-WORLD CONNECTION
Think of a warehouse conveyor belt where items enter from both ends and are paired for packaging; you only need to track the front and back positions, not the entire inventory, to form each package efficiently.
When you see a problem that mentions "first and last", "pair from both ends", or "move inward", immediately reach for the two‑pointer scaffold—declare left=0, right=n‑1, and loop while left<=right.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The pairing‑from‑both‑ends pattern is a classic example of two‑pointer technique on a linear data structure. By maintaining one pointer at the start of the array and another at the end, we can process complementary elements in a single linear scan, guaranteeing O(n) time while using O(1) extra space. This approach leverages the inherent order of the array and avoids the need for auxiliary data structures such as stacks or queues.
A naive solution would iterate over the array multiple times—first to compute the total sum, then to recompute pairwise sums, or even use nested loops to pair each element with every other. Such strategies explode to O(n²) time for large inputs and quickly exceed time limits. The optimal paradigm—two‑pointer traversal—collapses the problem to a single pass, because each element is visited exactly once, either as a left‑hand or right‑hand partner, and the algorithm terminates when the pointers cross.
The key insight is that the pairing rule defines a deterministic mapping: element i pairs with element n‑1‑i. Recognizing this bijection allows us to replace any recursive or index‑reversal logic with a simple while loop. This reduces both runtime and memory overhead, making the solution scalable to arrays with millions of elements.
Interview Questions on This Problem
Q1How would you modify the two‑pointer solution to also return the maximum pair sum while computing the total?
Maintain an additional variable maxPair initialized to negative infinity. Inside the while loop, compute the current pair sum (arr[left] + arr[right]) and update maxPair = max(maxPair, currentPair). After the loop, return both the total sum and maxPair.
Q2If the array is presented as a stream (you cannot index arbitrarily), can you still compute the required sum in O(1) extra space?
No. The pairing requires knowledge of the element at the symmetric position from the end, which is unavailable in a single‑pass stream without buffering. You would need to store the entire stream (or at least half of it) to later retrieve the counterpart, leading to O(n) space.
Q3Explain how the two‑pointer pattern for this problem relates to the "reverse‑pair" problem often asked in fintech coding interviews.
Both problems involve processing elements that are symmetric with respect to the array’s midpoint. In reverse‑pair questions you count pairs (i, j) where i < j and arr[i] > 2*arr[j]; the two‑pointer technique can be adapted by sorting and then scanning from both ends to efficiently count qualifying pairs, mirroring the way we pair first‑last elements here.
Examples
Input
nums = [1, 2, 3, 4, 5]
Output
15
Explanation: Step 1: Pair the first and last elements: 1 + 5 = 6. Step 2: Pair the second and second-to-last elements: 2 + 4 = 6. Step 3: The middle element is 3, which is added as a standalone term. Total sum = 6 + 6 + 3 = 15.
Input
nums = [10, 20, 30, 40]
Output
100
Explanation: Step 1: Pair the first and last elements: 10 + 40 = 50. Step 2: Pair the second and second-to-last elements: 20 + 30 = 50. Total sum = 50 + 50 = 100.
Input
nums = [7, 8, 9]
Output
24
Explanation: Step 1: Pair the first and last elements: 7 + 9 = 16. Step 2: The middle element is 8, which is added as a standalone term. Total sum = 16 + 8 = 24.
Input
nums = [1, 1, 1, 1, 1, 1]
Output
6
Explanation: Step 1: Pair the first and last elements: 1 + 1 = 2. Step 2: Pair the second and second-to-last elements: 1 + 1 = 2. Step 3: Pair the third and third-to-last elements: 1 + 1 = 2. Total sum = 2 + 2 + 2 = 6.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The array must contain at least one element.
- The sum of all elements must fit within a 64-bit integer.
Optimal Approach & Strategy
Apply a two‑pointer scan from both ends, adding each pair in a single linear pass, achieving O(n) time and O(1) space.
Brute Force Approach
Use a nested loop or separate passes to first compute the total sum, then recompute pairwise sums, resulting in O(n²) time.
Verified Code Solutions
function sumArray(nums) {
let sum = 0;
for (let num of nums) sum += num;
return sum;
}#include <bits/stdc++.h>
using namespace std;
int sumArray(const vector<int>& nums) {
long long sum = 0;
for (int num : nums) sum += num;
return sum;
}
public class Solution {
public int sumArray(int[] nums) {
int sum = 0;
for (int num : nums) sum += num;
return sum;
}
}
def sum_array(nums):
return sum(nums)
function sumArray(nums) {
let sum = 0;
for (let num of nums) sum += num;
return sum;
}
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