Alternating Array Reconstruction — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length n that stores the successive differences between adjacent elements of a hidden array orig. The hidden array always starts with orig[0]=0. For each index i (0‑based) the difference nums[i] is applied to the current value of orig[i] as follows: if i is even, orig[i+1]=orig[i]+nums[i]; if i is odd, orig[i+1]=orig[i]-nums[i]. Construct and return the complete array orig of length n+1. The algorithm must run in linear time.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Alternating Array Reconstruction"
WHY DOES IT MATTER?
This pattern is essential for problems involving cumulative changes, such as stock price reconstruction from daily changes or signal processing. It teaches the importance of recognizing linear dependencies and avoiding unnecessary complexity.
OPTIMIZATION CHALLENGE
The key insight is that the reconstruction is a single-pass operation. There is no need for sorting, hashing, or dynamic programming; a simple loop with a running sum suffices.
REAL-WORLD CONNECTION
Analogous to reconstructing a GPS trajectory from a series of velocity and direction changes, where each step depends on the previous position and the current movement vector.
In interviews, explicitly state that you are using a 'running sum' or 'prefix sum' approach. This demonstrates algorithmic awareness and helps the interviewer follow your logic.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem relies on the concept of prefix sums with a conditional sign flip, often referred to as an alternating prefix sum. The hidden array orig is reconstructed by iterating through the difference array nums and accumulating the values. The key theoretical insight is that the operation is deterministic and linear: each element in orig depends only on the previous element and the current difference in nums. This transforms the problem from a complex reconstruction task into a simple linear scan, avoiding any need for backtracking or complex data structures.
Interview Questions on This Problem
Q1How would you modify this algorithm if the sign flip depended on the value of `nums[i]` rather than the index `i`?
You would replace the i % 2 check with a condition on nums[i] (e.g., if nums[i] > 0). The time complexity remains O(n), but you must ensure the logic for adding or subtracting is correctly mapped to the new condition.
Q2What is the space complexity of this solution, and can it be optimized further?
The space complexity is O(1) if you modify the input array or use a single variable to track the current value, excluding the output array. If the output array is required, the space is O(n) for the result, but no auxiliary data structures are needed.
Q3How would you handle integer overflow in this problem?
Use a 64-bit integer (long) for the accumulator variable to prevent overflow during the summation process, especially if the input values are large or the array is long.
Examples
Input
[4,1,3]
Output
[0,4,3,6]
Explanation: Start with 0. i=0 (even): 0+4=4 → orig[1]=4. i=1 (odd): 4-1=3 → orig[2]=3. i=2 (even): 3+3=6 → orig[3]=6. Final array: [0,4,3,6].
Input
[-2,5,-1,2]
Output
[0,-2,-7,-8,-10]
Explanation: orig[0]=0. i=0 (even): 0+(-2)=-2. i=1 (odd): -2-5=-7. i=2 (even): -7+(-1)=-8. i=3 (odd): -8-2=-10. Resulting array is [0,-2,-7,-8,-10].
Input
[10]
Output
[0,10]
Explanation: Only one difference. i=0 (even): 0+10=10 → orig[1]=10. Output is [0,10].
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- All calculations fit in 64‑bit signed integer range
Optimal Approach & Strategy
Use a single loop to iterate through the input array, maintaining a running sum. For each index, add or subtract the current element based on whether the index is even or odd, and store the result in the output array.
Brute Force Approach
A naive approach might involve recursively trying all possible sign combinations, which is exponential and unnecessary. Another naive approach could be to store all intermediate states in a list, which is inefficient in space.
Verified Code Solutions
function reconstruct(nums) {
const orig = [];
let cur = 0;
orig.push(cur);
for (let i = 0; i < nums.length; ++i) {
if (i % 2 === 0) cur += nums[i];
else cur -= nums[i];
orig.push(cur);
}
return orig;
}
function main() {
const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim();
if (data.length === 0) {
console.log('0');
return;
}
const nums = data.split(/\s+/).map(Number);
const orig = reconstruct(nums);
console.log(orig.join(' '));
}
main();#include <bits/stdc++.h>
using namespace std;
vector<long long> reconstruct(const vector<long long>& nums) {
vector<long long> orig;
orig.reserve(nums.size() + 1);
long long cur = 0;
orig.push_back(cur);
for (size_t i = 0; i < nums.size(); ++i) {
if (i % 2 == 0) cur += nums[i];
else cur -= nums[i];
orig.push_back(cur);
}
return orig;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
vector<long long> nums;
long long x;
while (cin >> x) nums.push_back(x);
vector<long long> orig = reconstruct(nums);
for (size_t i = 0; i < orig.size(); ++i) {
if (i) cout << ' ';
cout << orig[i];
}
cout << '\n';
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static List<Long> reconstruct(List<Long> nums) {
List<Long> orig = new ArrayList<>(nums.size() + 1);
long cur = 0L;
orig.add(cur);
for (int i = 0; i < nums.size(); i++) {
long v = nums.get(i);
if (i % 2 == 0) cur += v;
else cur -= v;
orig.add(cur);
}
return orig;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
List<Long> nums = new ArrayList<>();
while ((line = br.readLine()) != null) {
line = line.trim();
if (line.isEmpty()) continue;
for (String s : line.split("\\s+")) {
nums.add(Long.parseLong(s));
}
}
List<Long> orig = reconstruct(nums);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < orig.size(); i++) {
if (i > 0) sb.append(' ');
sb.append(orig.get(i));
}
System.out.println(sb.toString());
}
}
def reconstruct(nums):
orig = [0]
cur = 0
for i, v in enumerate(nums):
if i % 2 == 0:
cur += v
else:
cur -= v
orig.append(cur)
return orig
if __name__ == "__main__":
import sys
data = sys.stdin.read().strip().split()
nums = list(map(int, data))
orig = reconstruct(nums)
print(' '.join(map(str, orig)))
function reconstruct(nums) {
const orig = [];
let cur = 0;
orig.push(cur);
for (let i = 0; i < nums.length; ++i) {
if (i % 2 === 0) cur += nums[i];
else cur -= nums[i];
orig.push(cur);
}
return orig;
}
function main() {
const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim();
if (data.length === 0) {
console.log('0');
return;
}
const nums = data.split(/\s+/).map(Number);
const orig = reconstruct(nums);
console.log(orig.join(' '));
}
main();
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