Maximum Profitable Triplets With Increasing Prices I solution and explanation
IBM
The problem
Given the 0-indexed arrays prices and profits of length n. There are n items in an store where the ith item has a price of prices[i] and a profit of profits[i].
We have to pick three items with the following condition:
prices[i] < prices[j] < prices[k]wherei < j < k.
If we pick items with indices i, j and k satisfying the above condition, the profit would be profits[i] + profits[j] + profits[k].
Return the maximum profit we can get, and -1 if it's not possible to pick three items with the given condition.
Example 1:
Input: prices = [10,2,3,4], profits = [100,2,7,10] Output: 19 Explanation: We can't pick the item with index i=0 since there are no indices j and k such that the condition holds. So the only triplet we can pick, are the items with indices 1, 2 and 3 and it's a valid pick since prices[1] < prices[2] < prices[3]. The answer would be sum of their profits which is 2 + 7 + 10 = 19.
Example 2:
Input: prices = [1,2,3,4,5], profits = [1,5,3,4,6] Output: 15 Explanation: We can select any triplet of items since for each triplet of indices i, j and k such that i < j < k, the condition holds. Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4. The answer would be sum of their profits which is 5 + 4 + 6 = 15.
Example 3:
Input: prices = [4,3,2,1], profits = [33,20,19,87] Output: -1 Explanation: We can't select any triplet of indices such that the condition holds, so we return -1.
Constraints:
3 <= prices.length == profits.length <= 20001 <= prices[i] <= 1061 <= profits[i] <= 106
Approach overview
Problem Overview: You are given two arrays prices and profits. Choose three indices i < j < k such that prices[i] < prices[j] < prices[k]. Among all valid triplets, return the maximum possible sum profits[i] + profits[j] + profits[k]. If no valid triplet exists, return -1.
Approach 1: Enumerate the Middle Element (O(n²) time, O(1) space)
Treat each index j as the middle element of the triplet. Scan the left side to find the maximum profits[i] where prices[i] < prices[j]. Then scan the right side to find the maximum profits[k] where prices[k] > prices[j]. If both exist, compute the candidate profit profits[i] + profits[j] + profits[k] and update the answer. This brute-force style enumeration works because every valid triplet must have a unique middle element. The downside is the nested scanning: for each j, you potentially traverse the entire prefix and suffix, leading to O(n²) time and O(1) extra space. Still useful as a baseline and easy to reason about.
Approach 2: Binary Indexed Tree (O(n log n) time, O(n) space)
The expensive part of the previous approach is repeatedly searching for the best profit on the left and right with price constraints. A Binary Indexed Tree (Fenwick Tree) stores the maximum profit seen so far for compressed price values and answers these queries in O(log n). First coordinate-compress the values in prices. Sweep from left to right and use the BIT to query the maximum profit for any price smaller than prices[j]. Store this as the best left candidate. Then sweep from right to left using another BIT to compute the best right candidate where price is greater than the current one. Once both arrays are built, combine them with profits[j] to evaluate each triplet in linear time. This reduces repeated scans and brings the complexity down to O(n log n) time with O(n) extra space.
This problem mixes ordering constraints with range maximum queries, which is why data structures from the segment tree or Fenwick tree family fit naturally. The input itself is a simple array, but the increasing price condition turns it into a range query problem.
Recommended for interviews: Start by explaining the middle-element enumeration. It proves you understand the structure of the triplet constraint. Then move to the Binary Indexed Tree optimization that answers "best profit with smaller/greater price" queries in O(log n). Interviewers usually expect the O(n log n) approach because it demonstrates familiarity with coordinate compression and Fenwick Trees for range maximum queries.
Approach 1: Enumerate the Middle Element
We can enumerate the middle element profits[j], and then enumerate the left element profits[i] and the right element profits[k]. For each profits[j], we need to find the maximum profits[i] and the maximum profits[k] such that prices[i] < prices[j] < prices[k]. We define left as the maximum value on the left of profits[j], and right as the maximum value on the right of profits[j]. If they exist, we update the answer as ans = max(ans, left + profits[j] + right).
The time complexity is O(n^2), where n is the length of the array. The space complexity is O(1).
class Solution:
def maxProfit(self, prices: List[int], profits: List[int]) -> int:
n = len(prices)
ans = -1
for j, x in enumerate(profits):
left = right = 0
for i in range(j):
if prices[i] < prices[j] and left < profits[i]:
left = profits[i]
for k in range(j + 1, n):
if prices[j] < prices[k] and right < profits[k]:
right = profits[k]
if left and right:
ans = max(ans, left + x + right)
return ansclass Solution {
public int maxProfit(int[] prices, int[] profits) {
int n = prices.length;
int ans = -1;
for (int j = 0; j < n; ++j) {
int left = 0, right = 0;
for (int i = 0; i < j; ++i) {
if (prices[i] < prices[j]) {
left = Math.max(left, profits[i]);
}
}
for (int k = j + 1; k < n; ++k) {
if (prices[j] < prices[k]) {
right = Math.max(right, profits[k]);
}
}
if (left > 0 && right > 0) {
ans = Math.max(ans, left + profits[j] + right);
}
}
return ans;
}
}class Solution {
public:
int maxProfit(vector<int>& prices, vector<int>& profits) {
int n = prices.size();
int ans = -1;
for (int j = 0; j < n; ++j) {
int left = 0, right = 0;
for (int i = 0; i < j; ++i) {
if (prices[i] < prices[j]) {
left = max(left, profits[i]);
}
}
for (int k = j + 1; k < n; ++k) {
if (prices[j] < prices[k]) {
right = max(right, profits[k]);
}
}
if (left && right) {
ans = max(ans, left + profits[j] + right);
}
}
return ans;
}
};function maxProfit(prices: number[], profits: number[]): number {
const n = prices.length;
let ans = -1;
for (let j = 0; j < n; ++j) {
let [left, right] = [0, 0];
for (let i = 0; i < j; ++i) {
if (prices[i] < prices[j]) {
left = Math.max(left, profits[i]);
}
}
for (let k = j + 1; k < n; ++k) {
if (prices[j] < prices[k]) {
right = Math.max(right, profits[k]);
}
}
if (left > 0 && right > 0) {
ans = Math.max(ans, left + profits[j] + right);
}
}
return ans;
}func maxProfit(prices []int, profits []int) int {
n := len(prices)
ans := -1
for j, x := range profits {
left, right := 0, 0
for i := 0; i < j; i++ {
if prices[i] < prices[j] {
left = max(left, profits[i])
}
}
for k := j + 1; k < n; k++ {
if prices[j] < prices[k] {
right = max(right, profits[k])
}
}
if left > 0 && right > 0 {
ans = max(ans, left+x+right)
}
}
return ans
}impl Solution {
pub fn max_profit(prices: Vec<i32>, profits: Vec<i32>) -> i32 {
let n = prices.len();
let mut ans = -1;
for j in 0..n {
let mut left = 0;
let mut right = 0;
for i in 0..j {
if prices[i] < prices[j] {
left = left.max(profits[i]);
}
}
for k in j + 1..n {
if prices[j] < prices[k] {
right = right.max(profits[k]);
}
}
if left > 0 && right > 0 {
ans = ans.max(left + profits[j] + right);
}
}
ans
}
}Approach 2: Binary Indexed Tree
We can use two Binary Indexed Trees (BITs) to maintain the maximum profit on the left and right of each price, respectively. Then, we enumerate the middle price, query the maximum profit on both sides through the BIT, and finally take the maximum value.
The time complexity is O(n times log M), and the space complexity is O(M). Here, n is the length of the array prices, and M is the maximum value in the array prices. In this problem, M \le 10^6.
Since the length of prices does not exceed 2000, and the value of prices[i] reaches 10^6, we can discretize prices, which can reduce the space complexity to O(n).
class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = max(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mx = 0
while x:
mx = max(mx, self.c[x])
x -= x & -x
return mx
class Solution:
def maxProfit(self, prices: List[int], profits: List[int]) -> int:
n = len(prices)
left = [0] * n
right = [0] * n
m = max(prices)
tree1 = BinaryIndexedTree(m + 1)
tree2 = BinaryIndexedTree(m + 1)
for i, x in enumerate(prices):
left[i] = tree1.query(x - 1)
tree1.update(x, profits[i])
for i in range(n - 1, -1, -1):
x = m + 1 - prices[i]
right[i] = tree2.query(x - 1)
tree2.update(x, profits[i])
return max(
(l + x + r for l, x, r in zip(left, profits, right) if l and r), default=-1
)class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}
public void update(int x, int v) {
while (x <= n) {
c[x] = Math.max(c[x], v);
x += x & -x;
}
}
public int query(int x) {
int mx = 0;
while (x > 0) {
mx = Math.max(mx, c[x]);
x -= x & -x;
}
return mx;
}
}
class Solution {
public int maxProfit(int[] prices, int[] profits) {
int n = prices.length;
int[] left = new int[n];
int[] right = new int[n];
int m = 0;
for (int x : prices) {
m = Math.max(m, x);
}
BinaryIndexedTree tree1 = new BinaryIndexedTree(m + 1);
BinaryIndexedTree tree2 = new BinaryIndexedTree(m + 1);
for (int i = 0; i < n; ++i) {
int x = prices[i];
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (int i = n - 1; i >= 0; --i) {
int x = m + 1 - prices[i];
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
int ans = -1;
for (int i = 0; i < n; ++i) {
if (left[i] > 0 && right[i] > 0) {
ans = Math.max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}
}class BinaryIndexedTree {
private:
int n;
vector<int> c;
public:
BinaryIndexedTree(int n) {
this->n = n;
c.resize(n + 1, 0);
}
void update(int x, int v) {
while (x <= n) {
c[x] = max(c[x], v);
x += x & -x;
}
}
int query(int x) {
int mx = 0;
while (x > 0) {
mx = max(mx, c[x]);
x -= x & -x;
}
return mx;
}
};
class Solution {
public:
int maxProfit(vector<int>& prices, vector<int>& profits) {
int n = prices.size();
vector<int> left(n, 0);
vector<int> right(n, 0);
int m = *max_element(prices.begin(), prices.end());
BinaryIndexedTree tree1(m + 1);
BinaryIndexedTree tree2(m + 1);
for (int i = 0; i < n; ++i) {
int x = prices[i];
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (int i = n - 1; i >= 0; --i) {
int x = m + 1 - prices[i];
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
int ans = -1;
for (int i = 0; i < n; ++i) {
if (left[i] > 0 && right[i] > 0) {
ans = max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}
};class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}
update(x: number, v: number): void {
while (x <= this.n) {
this.c[x] = Math.max(this.c[x], v);
x += x & -x;
}
}
query(x: number): number {
let mx = 0;
while (x > 0) {
mx = Math.max(mx, this.c[x]);
x -= x & -x;
}
return mx;
}
}
function maxProfit(prices: number[], profits: number[]): number {
const n: number = prices.length;
const left: number[] = Array(n).fill(0);
const right: number[] = Array(n).fill(0);
const m = Math.max(...prices);
const tree1: BinaryIndexedTree = new BinaryIndexedTree(m + 1);
const tree2: BinaryIndexedTree = new BinaryIndexedTree(m + 1);
for (let i = 0; i < n; i++) {
const x: number = prices[i];
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (let i = n - 1; i >= 0; i--) {
const x: number = m + 1 - prices[i];
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
let ans: number = -1;
for (let i = 0; i < n; i++) {
if (left[i] > 0 && right[i] > 0) {
ans = Math.max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}type BinaryIndexedTree struct {
n int
c []int
}
func NewBinaryIndexedTree(n int) BinaryIndexedTree {
c := make([]int, n+1)
return BinaryIndexedTree{n: n, c: c}
}
func (bit *BinaryIndexedTree) update(x, v int) {
for x <= bit.n {
bit.c[x] = max(bit.c[x], v)
x += x & -x
}
}
func (bit *BinaryIndexedTree) query(x int) int {
mx := 0
for x > 0 {
mx = max(mx, bit.c[x])
x -= x & -x
}
return mx
}
func maxProfit(prices []int, profits []int) int {
n := len(prices)
left := make([]int, n)
right := make([]int, n)
m := slices.Max(prices)
tree1 := NewBinaryIndexedTree(m + 1)
tree2 := NewBinaryIndexedTree(m + 1)
for i, x := range prices {
left[i] = tree1.query(x - 1)
tree1.update(x, profits[i])
}
for i := n - 1; i >= 0; i-- {
x := m + 1 - prices[i]
right[i] = tree2.query(x - 1)
tree2.update(x, profits[i])
}
ans := -1
for i := 0; i < n; i++ {
if left[i] > 0 && right[i] > 0 {
ans = max(ans, left[i]+profits[i]+right[i])
}
}
return ans
}struct BinaryIndexedTree {
n: usize,
c: Vec<i32>,
}
impl BinaryIndexedTree {
fn new(n: usize) -> BinaryIndexedTree {
BinaryIndexedTree {
n,
c: vec![0; n + 1],
}
}
fn update(&mut self, x: usize, v: i32) {
let mut x = x;
while x <= self.n {
self.c[x] = self.c[x].max(v);
x += x & x.wrapping_neg();
}
}
fn query(&self, x: usize) -> i32 {
let mut x = x;
let mut mx = 0;
while x > 0 {
mx = mx.max(self.c[x]);
x -= x & x.wrapping_neg();
}
mx
}
}
impl Solution {
pub fn max_profit(prices: Vec<i32>, profits: Vec<i32>) -> i32 {
let n = prices.len();
let mut left = vec![0; n];
let mut right = vec![0; n];
let m = prices.iter().cloned().max().unwrap_or(0);
let mut tree1 = BinaryIndexedTree::new((m as usize) + 1);
let mut tree2 = BinaryIndexedTree::new((m as usize) + 1);
for i in 0..n {
let x = prices[i] as usize;
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for i in (0..n).rev() {
let x = (m + 1 - prices[i]) as usize;
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
let mut ans = -1;
for i in 0..n {
if left[i] > 0 && right[i] > 0 {
ans = ans.max(left[i] + profits[i] + right[i]);
}
}
ans
}
}Approach 3: Default Approach
class BinaryIndexedTree:
def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, v: int):
while x <= self.n:
self.c[x] = max(self.c[x], v)
x += x & -x
def query(self, x: int) -> int:
mx = 0
while x:
mx = max(mx, self.c[x])
x -= x & -x
return mx
class Solution:
def maxProfit(self, prices: List[int], profits: List[int]) -> int:
n = len(prices)
left = [0] * n
right = [0] * n
s = sorted(set(prices))
m = len(s)
tree1 = BinaryIndexedTree(m + 1)
tree2 = BinaryIndexedTree(m + 1)
for i, x in enumerate(prices):
x = bisect_left(s, x) + 1
left[i] = tree1.query(x - 1)
tree1.update(x, profits[i])
for i in range(n - 1, -1, -1):
x = m + 1 - (bisect_left(s, prices[i]) + 1)
right[i] = tree2.query(x - 1)
tree2.update(x, profits[i])
return max(
(l + x + r for l, x, r in zip(left, profits, right) if l and r), default=-1
)class BinaryIndexedTree {
private int n;
private int[] c;
public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
}
public void update(int x, int v) {
while (x <= n) {
c[x] = Math.max(c[x], v);
x += x & -x;
}
}
public int query(int x) {
int mx = 0;
while (x > 0) {
mx = Math.max(mx, c[x]);
x -= x & -x;
}
return mx;
}
}
class Solution {
public int maxProfit(int[] prices, int[] profits) {
int n = prices.length;
int[] left = new int[n];
int[] right = new int[n];
int[] s = prices.clone();
Arrays.sort(s);
int m = 0;
for (int i = 0; i < n; ++i) {
if (i == 0 || s[i] != s[i - 1]) {
s[m++] = s[i];
}
}
BinaryIndexedTree tree1 = new BinaryIndexedTree(m + 1);
BinaryIndexedTree tree2 = new BinaryIndexedTree(m + 1);
for (int i = 0; i < n; ++i) {
int x = search(s, prices[i], m);
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (int i = n - 1; i >= 0; --i) {
int x = m + 1 - search(s, prices[i], m);
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
int ans = -1;
for (int i = 0; i < n; ++i) {
if (left[i] > 0 && right[i] > 0) {
ans = Math.max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}
private int search(int[] nums, int x, int r) {
int l = 0;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l + 1;
}
}class BinaryIndexedTree {
private:
int n;
vector<int> c;
public:
BinaryIndexedTree(int n) {
this->n = n;
c.resize(n + 1, 0);
}
void update(int x, int v) {
while (x <= n) {
c[x] = max(c[x], v);
x += x & -x;
}
}
int query(int x) {
int mx = 0;
while (x > 0) {
mx = max(mx, c[x]);
x -= x & -x;
}
return mx;
}
};
class Solution {
public:
int maxProfit(vector<int>& prices, vector<int>& profits) {
int n = prices.size();
vector<int> left(n);
vector<int> right(n);
vector<int> s = prices;
sort(s.begin(), s.end());
s.erase(unique(s.begin(), s.end()), s.end());
int m = s.size();
BinaryIndexedTree tree1(m + 1);
BinaryIndexedTree tree2(m + 1);
for (int i = 0; i < n; ++i) {
int x = lower_bound(s.begin(), s.end(), prices[i]) - s.begin() + 1;
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (int i = n - 1; i >= 0; --i) {
int x = m + 1 - (lower_bound(s.begin(), s.end(), prices[i]) - s.begin() + 1);
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
int ans = -1;
for (int i = 0; i < n; ++i) {
if (left[i] > 0 && right[i] > 0) {
ans = max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}
};class BinaryIndexedTree {
private n: number;
private c: number[];
constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}
update(x: number, v: number): void {
while (x <= this.n) {
this.c[x] = Math.max(this.c[x], v);
x += x & -x;
}
}
query(x: number): number {
let mx = 0;
while (x > 0) {
mx = Math.max(mx, this.c[x]);
x -= x & -x;
}
return mx;
}
}
function maxProfit(prices: number[], profits: number[]): number {
const n: number = prices.length;
const left: number[] = Array(n).fill(0);
const right: number[] = Array(n).fill(0);
const s = [...prices].sort((a, b) => a - b);
let m = 0;
for (let i = 0; i < n; ++i) {
if (i === 0 || s[i] !== s[i - 1]) {
s[m++] = s[i];
}
}
s.length = m;
const tree1: BinaryIndexedTree = new BinaryIndexedTree(m + 1);
const tree2: BinaryIndexedTree = new BinaryIndexedTree(m + 1);
const search = (nums: number[], x: number): number => {
let [l, r] = [0, nums.length];
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] >= x) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
};
for (let i = 0; i < n; ++i) {
const x = search(s, prices[i]) + 1;
left[i] = tree1.query(x - 1);
tree1.update(x, profits[i]);
}
for (let i = n - 1; i >= 0; i--) {
const x: number = m + 1 - (search(s, prices[i]) + 1);
right[i] = tree2.query(x - 1);
tree2.update(x, profits[i]);
}
let ans: number = -1;
for (let i = 0; i < n; i++) {
if (left[i] > 0 && right[i] > 0) {
ans = Math.max(ans, left[i] + profits[i] + right[i]);
}
}
return ans;
}type BinaryIndexedTree struct {
n int
c []int
}
func NewBinaryIndexedTree(n int) BinaryIndexedTree {
c := make([]int, n+1)
return BinaryIndexedTree{n: n, c: c}
}
func (bit *BinaryIndexedTree) update(x, v int) {
for x <= bit.n {
bit.c[x] = max(bit.c[x], v)
x += x & -x
}
}
func (bit *BinaryIndexedTree) query(x int) int {
mx := 0
for x > 0 {
mx = max(mx, bit.c[x])
x -= x & -x
}
return mx
}
func maxProfit(prices []int, profits []int) int {
n := len(prices)
left := make([]int, n)
right := make([]int, n)
s := make([]int, n)
copy(s, prices)
sort.Ints(s)
m := 0
for i, x := range s {
if i == 0 || x != s[i-1] {
s[m] = x
m++
}
}
tree1 := NewBinaryIndexedTree(m + 1)
tree2 := NewBinaryIndexedTree(m + 1)
for i, x := range prices {
x = sort.SearchInts(s[:m], x) + 1
left[i] = tree1.query(x - 1)
tree1.update(x, profits[i])
}
for i := n - 1; i >= 0; i-- {
x := m + 1 - (sort.SearchInts(s[:m], prices[i]) + 1)
right[i] = tree2.query(x - 1)
tree2.update(x, profits[i])
}
ans := -1
for i := 0; i < n; i++ {
if left[i] > 0 && right[i] > 0 {
ans = max(ans, left[i]+profits[i]+right[i])
}
}
return ans
}Complexity comparison
| Approach | Complexity |
|---|---|
| Enumerate the Middle Element | – |
| Binary Indexed Tree | – |
| Default Approach | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Enumerate the Middle Element | O(n²) | O(1) | Good baseline solution and easy to implement when constraints are small. |
| Binary Indexed Tree (Fenwick Tree) | O(n log n) | O(n) | Best general solution for large inputs where repeated range maximum queries are needed. |
Frequently asked questions
Is Maximum Profitable Triplets With Increasing Prices I easy or hard?
This problem is generally rated Medium. The brute-force reasoning is straightforward, but reaching the optimal O(n log n) solution requires understanding Fenwick Trees or Segment Trees along with coordinate compression.
Maximum Profitable Triplets With Increasing Prices I Python/Java solution
Python and Java implementations usually follow the same pattern: coordinate compress the prices, maintain a Fenwick Tree storing maximum profit, and perform prefix queries to find the best candidate with a smaller price. A second pass from the right computes the best larger-price candidate before combining results.
How to solve Maximum Profitable Triplets With Increasing Prices I in O(n log n)?
Compress the price values and use a Binary Indexed Tree to store the maximum profit seen for each price rank. Sweep left to right to compute the best profit for a smaller price, then sweep right to left for larger prices. Combine these with the current profit to evaluate each possible middle element in linear time.
What is the best approach for Maximum Profitable Triplets With Increasing Prices I?
The most efficient approach uses a Binary Indexed Tree (Fenwick Tree) with coordinate compression. It allows querying the maximum profit for prices smaller or larger than the current element in O(log n). By computing the best left and right candidates for every index, the overall complexity becomes O(n log n) with O(n) space.
Is Maximum Profitable Triplets With Increasing Prices I asked at Google/Amazon/Meta?
Problems combining increasing constraints with range queries appear frequently in interviews at large tech companies such as Google, Amazon, and Meta. Variants often require Fenwick Trees, Segment Trees, or ordered maps to efficiently query prefix or suffix ranges.
What data structure is used in Maximum Profitable Triplets With Increasing Prices I?
The optimal solution typically uses a Binary Indexed Tree (Fenwick Tree) to maintain prefix maximum profits for compressed price indices. A Segment Tree can also solve the same range maximum query problem with similar O(n log n) complexity.
What is the time complexity of Maximum Profitable Triplets With Increasing Prices I?
A straightforward enumeration of the middle element takes O(n^2) time because each index scans both the prefix and suffix to find valid partners. The optimized approach using a Binary Indexed Tree reduces the complexity to O(n log n) by turning repeated scans into logarithmic range maximum queries.