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# Posts tagged as “range”

## Solution 1: Range sum query

Assuming we can collect fruits in range [l, r], we need a fast query to compute the sum of those fruits.

Given startPos and k, we have four options:
1. move i steps to the left
2. move i steps to the left and k – i steps to the right.
3. move i steps to the right
4. move i steps to the right and k – i steps to the left.

We enumerate i steps and calculate maximum range [l, r] covered by each option, and collect all the fruit in that range.

Time complexity: O(m + k)
Space complexity: O(m)
where m = max(max(pos), startPos)

## Solution 2: Sliding Window

Maintain a window [l, r] such that the steps to cover [l, r] from startPos is less or equal to k.

Time complexity: O(n)
Space complexity: O(1)

## C++

You are given a 2D integer array ranges and two integers left and right. Each ranges[i] = [starti, endi] represents an inclusive interval between starti and endi.

Return true if each integer in the inclusive range [left, right] is covered by at least one interval in ranges. Return false otherwise.

An integer x is covered by an interval ranges[i] = [starti, endi] if starti <= x <= endi.

Example 1:

Input: ranges = [[1,2],[3,4],[5,6]], left = 2, right = 5
Output: true
Explanation: Every integer between 2 and 5 is covered:
- 2 is covered by the first range.
- 3 and 4 are covered by the second range.
- 5 is covered by the third range.


Example 2:

Input: ranges = [[1,10],[10,20]], left = 21, right = 21
Output: false
Explanation: 21 is not covered by any range.


Constraints:

• 1 <= ranges.length <= 50
• 1 <= starti <= endi <= 50
• 1 <= left <= right <= 50

## Solution 1: Hashtable

Time complexity: O(n * (right – left))
Space complexity: O(right – left)

## C++

You are given an array of events where events[i] = [startDayi, endDayi, valuei]. The ith event starts at startDayiand ends at endDayi, and if you attend this event, you will receive a value of valuei. You are also given an integer k which represents the maximum number of events you can attend.

You can only attend one event at a time. If you choose to attend an event, you must attend the entire event. Note that the end day is inclusive: that is, you cannot attend two events where one of them starts and the other ends on the same day.

Return the maximum sum of values that you can receive by attending events.

Example 1:

Input: events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
Output: 7
Explanation: Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.

Example 2:

Input: events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
Output: 10
Explanation: Choose event 2 for a total value of 10.
Notice that you cannot attend any other event as they overlap, and that you do not have to attend k events.

Example 3:

Input: events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
Output: 9
Explanation: Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.

Constraints:

• 1 <= k <= events.length
• 1 <= k * events.length <= 106
• 1 <= startDayi <= endDayi <= 109
• 1 <= valuei <= 106

## Solution: DP + Binary Search

Sort events by ending time.
dp[i][j] := max value we can get by attending at most j events among events[0~i].
dp[i][j] = max(dp[i – 1][j], dp[p][j – 1] + value[i])
p is the first event that does not overlap with the current one.

Time complexity: O(nlogn + nk)
Space complexity: O(nk)

## C++

There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.

In each round of the game, Alice divides the row into two non-empty rows (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the values of all the stones in this row. Bob throws away the row which has the maximum value, and Alice’s score increases by the value of the remaining row. If the value of the two rows are equal, Bob lets Alice decide which row will be thrown away. The next round starts with the remaining row.

The game ends when there is only one stone remaining. Alice’s is initially zero.

Return the maximum score that Alice can obtain.

Example 1:

Input: stoneValue = [6,2,3,4,5,5]
Output: 18
Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11.
In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5).
The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row.


Example 2:

Input: stoneValue = [7,7,7,7,7,7,7]
Output: 28


Example 3:

Input: stoneValue = [4]
Output: 0


Constraints:

• 1 <= stoneValue.length <= 500
• 1 <= stoneValue[i] <= 10^6

## Solution: Range DP + Prefix Sum

dp[l][r] := max store Alice can get from range [l, r]
sum_l = sum(l, k), sum_r = sum(k + 1, r)
dp[l][r] = max{
dp[l][k] + sum_l if sum_l < sum_r
dp[k+1][r] + sum_r if sum_r < sum_l
max(dp[l][k], dp[k+1][r])) + sum_l if sum_l == sum_r)
} for k in [l, r)

Time complexity: O(n^3)
Space complexity: O(n^2)

# Problem

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

Example 1:

Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32


Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23


Note:

1. The number of nodes in the tree is at most 10000.
2. The final answer is guaranteed to be less than 2^31.

# Solution: In-order traversal

Time complexity: O(n)

Space complexity: O(n)