Alternate Tree Fruiting — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Alternate Tree Fruiting problem optimally.
O(n)O(1)Problem Description
Given an integer array treeFruits of length n, select a non‑empty subsequence of elements whose indices form an arithmetic progression with common difference 2. In other words, if the first chosen index is i, the next must be i+2, then i+4, and so on, staying within the array bounds. The subsequence may start at any valid index and may end at any point; you may also choose a single element. Return the maximum possible sum of the selected elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Alternate Tree Fruiting"
WHY DOES IT MATTER?
Recognising that a fixed stride creates independent parity streams lets you convert a seemingly two‑dimensional combinatorial search into a classic one‑dimensional maximum subarray problem, a pattern that appears in many “skip‑step” or “alternating” constraints.
OPTIMIZATION CHALLENGE
The key insight is that the progression never jumps between parity groups, so you can collapse the problem to two simple sub‑problems and apply Kadane’s O(n) algorithm, cutting the naïve O(n^2) search down to linear time.
REAL-WORLD CONNECTION
Think of a distributed log replicated on two shards where reads alternate between shards; optimizing read‑through latency is equivalent to picking the best contiguous segment on a single shard.
During an interview, first state the parity observation, then immediately propose Kadane on each parity – this shows you can map a custom constraint to a well‑known algorithmic primitive.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to finding the maximum‑sum subsequence whose indices share the same parity because a fixed step of 2 never changes parity. Naïve enumeration of all possible start positions and lengths would be O(n^2) – each start would scan forward accumulating sums, which quickly blows up for n up to 10^5 or more. The optimal paradigm treats the even‑indexed elements as one independent array and the odd‑indexed elements as another, then applies Kadane’s linear‑time maximum subarray algorithm on each. This works because any valid progression is exactly a contiguous sub‑array in one of those parity‑filtered sequences, and Kadane guarantees the best contiguous sum in O(length) time. The final answer is the larger of the two Kadane results, yielding a linear‑time, constant‑space solution.
Interview Questions on This Problem
Q1How would you modify the solution if the common difference were 3 instead of 2?
Group the original array into three parity classes based on index mod 3, run Kadane on each class, and take the maximum; the time remains O(n) and space O(1).
Q2Can you compute the answer in a single pass without storing the two filtered arrays?
Yes – maintain two running Kadane states (current and best) for even and odd positions while iterating the original array, updating the appropriate state based on index parity.
Q3What edge case must you handle when all numbers are negative?
Kadane’s classic formulation returns the maximum element (the least negative) rather than zero, so you must initialise the best sum with the first element of each parity class and never reset to zero.
Examples
Input
[4,-1,2,5,-3,7]
Output
12
Explanation: Starting at index 3 (value 5) and then taking index 5 (value 7) yields the sum 5+7=12, which is larger than any other alternating‑step subsequence.
Input
[-5,-2,-3,-4]
Output
-2
Explanation: All alternating‑step subsequences contain negative numbers. The greatest sum is obtained by selecting the single element at index 1 (value -2).
Input
[10,1,10,1,10]
Output
30
Explanation: Choosing indices 0,2,4 gives 10+10+10=30, which is the maximum achievable sum.
Constraints
- 1 <= treeFruits.length <= 100000
- -1000000000 <= treeFruits[i] <= 1000000000
- The algorithm must run in O(n) time and O(1) additional space
Optimal Approach & Strategy
Separate even and odd indices, run Kadane’s algorithm on each in a single pass, and take the maximum – O(n) time, O(1) space.
Brute Force Approach
Enumerate every possible start index, then keep adding every second element while tracking the best sum – O(n^2) time.
Code Solutions
function maxFruitSum(treeFruits){
if(treeFruits.length===0) return 0;
const even=[], odd=[];
for(let i=0;i<treeFruits.length;i++){
if(i%2===0) even.push(treeFruits[i]);
else odd.push(treeFruits[i]);
}
const kadane = arr=>{
let best=arr[0], cur=arr[0];
for(let i=1;i<arr.length;i++){
cur = Math.max(arr[i], cur+arr[i]);
best = Math.max(best, cur);
}
return best;
};
let ans = -Infinity;
if(even.length) ans = Math.max(ans, kadane(even));
if(odd.length) ans = Math.max(ans, kadane(odd));
return ans;
}
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(input.length===0) process.exit(0);
let idx=0; const n=input[idx++]; const arr=input.slice(idx, idx+n);
console.log(maxFruitSum(arr));#include <bits/stdc++.h>
using namespace std;
long long kadane(const vector<int>& v){
long long best = v[0];
long long cur = v[0];
for(size_t i=1;i<v.size();++i){
cur = max<long long>(v[i], cur+v[i]);
best = max(best, cur);
}
return best;
}
long long maxFruitSum(const vector<int>& treeFruits){
if(treeFruits.empty()) return 0;
vector<int> even, odd;
for(size_t i=0;i<treeFruits.size();++i){
if(i%2==0) even.push_back(treeFruits[i]);
else odd.push_back(treeFruits[i]);
}
long long ans = LLONG_MIN;
if(!even.empty()) ans = max(ans, kadane(even));
if(!odd.empty()) ans = max(ans, kadane(odd));
return ans;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0; vector<int> a(n); for(int i=0;i<n;++i)cin>>a[i];
cout<<maxFruitSum(a);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
private static long kadane(int[] arr){
long best = arr[0];
long cur = arr[0];
for(int i=1;i<arr.length;i++){
cur = Math.max(arr[i], cur + arr[i]);
best = Math.max(best, cur);
}
return best;
}
public static long maxFruitSum(int[] treeFruits){
if(treeFruits.length==0) return 0;
int evenCount = (treeFruits.length+1)/2;
int oddCount = treeFruits.length/2;
int[] even = new int[evenCount];
int[] odd = new int[oddCount];
int ei=0, oi=0;
for(int i=0;i<treeFruits.length;i++){
if(i%2==0) even[ei++]=treeFruits[i];
else odd[oi++]=treeFruits[i];
}
long ans = Long.MIN_VALUE;
if(even.length>0) ans = Math.max(ans, kadane(even));
if(odd.length>0) ans = Math.max(ans, kadane(odd));
return ans;
}
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if(line==null||line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
int[] arr = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) arr[i]=Integer.parseInt(st.nextToken());
System.out.print(maxFruitSum(arr));
}
}def max_fruit_sum(treeFruits):
if not treeFruits:
return 0
even = treeFruits[0::2]
odd = treeFruits[1::2]
def kadane(arr):
best = cur = arr[0]
for x in arr[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
ans = max(kadane(even), kadane(odd) if odd else float('-inf'))
return ans
if __name__ == "__main__":
import sys
data = sys.stdin.read().strip().split()
if not data:
sys.exit(0)
it = iter(data)
n = int(next(it))
arr = [int(next(it)) for _ in range(n)]
print(max_fruit_sum(arr))function maxFruitSum(treeFruits){
if(treeFruits.length===0) return 0;
const even=[], odd=[];
for(let i=0;i<treeFruits.length;i++){
if(i%2===0) even.push(treeFruits[i]);
else odd.push(treeFruits[i]);
}
const kadane = arr=>{
let best=arr[0], cur=arr[0];
for(let i=1;i<arr.length;i++){
cur = Math.max(arr[i], cur+arr[i]);
best = Math.max(best, cur);
}
return best;
};
let ans = -Infinity;
if(even.length) ans = Math.max(ans, kadane(even));
if(odd.length) ans = Math.max(ans, kadane(odd));
return ans;
}
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(input.length===0) process.exit(0);
let idx=0; const n=input[idx++]; const arr=input.slice(idx, idx+n);
console.log(maxFruitSum(arr));Asked in Top Tech Interviews
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