Leetcode 2212: Maximum Points in an Archery Competition
Alice and Bob are opponents in an archery competition. Alice first shoots numArrows arrows, followed by Bob, in the target scoring sections from 0 to 11. The goal is to maximize Bob’s total score, ensuring that the sum of arrows shot by Bob equals numArrows.
Problem
Approach
Steps
Complexity
Input: You are given the total number of arrows Bob can shoot and an array representing the arrows shot by Alice in each section of the target. Bob's goal is to maximize his score while shooting exactly numArrows arrows.
Example: numArrows = 6, aliceArrows = [2, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0]
Constraints:
• 1 <= numArrows <= 100000
• aliceArrows.length == bobArrows.length == 12
• 0 <= aliceArrows[i], bobArrows[i] <= numArrows
• sum(aliceArrows[i]) == numArrows
Output: Return an array of length 12 representing the number of arrows Bob shoots in each scoring section. The sum of the values in this array must equal numArrows.
Example: [0, 1, 0, 0, 1, 1, 1, 0, 2, 0, 0, 0]
Constraints:
• The array must have exactly 12 elements.
Goal: Maximize Bob's score by strategically choosing the number of arrows to shoot in each section.
Steps:
• Iterate through all sections and decide where Bob can shoot more arrows than Alice.
• Calculate the total score for Bob based on these decisions and ensure Bob shoots exactly numArrows arrows.
Goal: The constraints ensure that the problem's input is manageable and can be processed efficiently.
Steps:
• The total number of arrows for Alice will always be numArrows.
Assumptions:
• Alice's arrow distribution is fixed.
• Bob can shoot any number of arrows, provided it sums to numArrows.
• Input: numArrows = 6, aliceArrows = [2, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0]
• Explanation: In this example, Alice shoots arrows in certain sections. Bob needs to shoot arrows in other sections where he can score points, ensuring his total shots equal numArrows.
Approach: The approach involves selecting sections where Bob can score maximum points and decide the number of arrows to shoot to achieve this.
Observations:
• Alice's arrows are fixed, so we can only adjust Bob's arrows in response.
• Bob needs to maximize the sections where he can take points.
• This problem involves a dynamic approach where we need to explore how Bob can beat Alice's arrows in each section while managing his total arrows.
Steps:
• 1. Use dynamic programming to decide which sections Bob should target.
• 2. Ensure that Bob's total arrows do not exceed numArrows.
• 3. Calculate the final result based on Bob's decisions.
Empty Inputs:
• Handle edge cases where no arrows are shot.
Large Inputs:
• Ensure the solution works for large values of numArrows.
Special Values:
• Consider edge cases like Alice shooting all arrows in one section.
Constraints:
• Optimize for large inputs to avoid time-limit exceed errors.
vector<int> bob, alice;
int n;
int mask;
int ans;
void dp(int idx, int rm, int pts, int msk) {
// cout << idx << " " << rm << " " << pts << "\n";
if((rm <= 0) || (idx == n)){
if(ans < pts) {
ans = pts;
mask = msk;
}
return;
}
dp(idx + 1, rm, pts, msk);
if(rm > alice[idx]) {
msk |= (1 << idx);
dp(idx + 1, rm - alice[idx] - 1, pts + idx, msk);
}
}
vector<int> maximumBobPoints(int net, vector<int>& alice) {
n = alice.size();
this->alice = alice;
bob.resize(n, 0);
int idx = 0;
int pts = 0;
mask = 0;
ans = 0;
dp(idx, net, pts, 0);
for(int i = n - 1; i >= 0; i--) {
if((mask >> i) & 1) {
bob[i] = alice[i] + 1;
net -= bob[i];
}
}
if(net > 0) bob[0] += net;
return bob;
}
1 : Variable Initialization
vector<int> bob, alice;
These are two vectors: 'bob' holds the points for Bob, and 'alice' holds the points for Alice.
2 : Variable Initialization
int n;
This integer variable 'n' stores the size of the 'alice' vector, which determines the number of rounds or problems.
3 : Variable Initialization
int mask;
The 'mask' integer holds a bitmask representing which problems Bob decides to attempt.
4 : Variable Initialization
int ans;
This integer variable stores the maximum points Bob can achieve.
5 : Helper Function Declaration
This section sets up the helper function for dynamic programming (dp) to calculate the optimal points for Bob.
6 : Helper Function Implementation
void dp(int idx, int rm, int pts, int msk) {
The 'dp' function is a recursive function that explores all the possibilities for Bob's points based on the current problem index, remaining rounds, and the current points and bitmask.
7 : Conditional Check
if((rm <= 0) || (idx == n)){
This checks if the remaining rounds are exhausted or if all problems have been considered.
8 : Conditional Block
if(ans < pts) {
If the current points exceed the maximum points found so far, update the answer.
9 : Assignment
ans = pts;
Assign the current points to 'ans' if they represent the new maximum.
10 : Assignment
mask = msk;
Update the bitmask with the current bitmask (msk) if this configuration yields more points.
11 : Return Statement
return;
Exit the current recursive call if the condition above is met.
12 : Condition Check
dp(idx + 1, rm, pts, msk);
Continue to the next problem without assigning it to Bob.
13 : Conditional Check
if(rm > alice[idx]) {
Check if Bob has enough remaining rounds to attempt the current problem.
14 : Bitmask Manipulation
msk |= (1 << idx);
Set the corresponding bit in the bitmask to represent that Bob attempts the current problem.
15 : Recursive Call
dp(idx + 1, rm - alice[idx] - 1, pts + idx, msk);
Recursive call to consider the next problem, updating remaining rounds and points.
16 : Main Function Definition
vector<int> maximumBobPoints(int net, vector<int>& alice) {
This function calculates Bob's maximum points given the total remaining points (net) and Alice's points for each problem.
17 : Variable Assignment
n = alice.size();
Assign the size of Alice's vector to 'n'.
18 : Variable Assignment
this->alice = alice;
Assign the passed-in 'alice' vector to the instance variable.
19 : Vector Initialization
bob.resize(n, 0);
Resize the 'bob' vector to match the size of Alice's vector and initialize all elements to 0.
20 : Variable Initialization
int idx = 0;
Initialize the index variable 'idx' to 0.
21 : Variable Initialization
int pts = 0;
Initialize the points variable 'pts' to 0.
22 : Variable Initialization
mask = 0;
Initialize the bitmask variable 'mask' to 0.
23 : Variable Initialization
ans = 0;
Initialize the answer variable 'ans' to 0.
24 : Function Call
dp(idx, net, pts, 0);
Call the dynamic programming function to compute the maximum points and the optimal configuration for Bob.
25 : Loop
for(int i = n - 1; i >= 0; i--) {
Loop through the problems in reverse order to assign the optimal values to the 'bob' vector.
26 : Condition Check
if((mask >> i) & 1) {
Check if the current bit in the bitmask indicates that Bob attempts this problem.
27 : Assignment
bob[i] = alice[i] + 1;
Assign the optimal points for Bob on this problem based on the bitmask.
28 : Variable Update
net -= bob[i];
Update the remaining points after Bob takes this problem.
29 : Condition Check
if(net > 0) bob[0] += net;
If there are remaining points, assign them to Bob's first problem.
30 : Return Statement
return bob;
Return the final 'bob' vector with the optimal points for each problem.
Best Case: O(numArrows)
Average Case: O(numArrows)
Worst Case: O(numArrows)
Description: The time complexity depends on how we process each section and the number of arrows Bob needs to shoot.
Best Case: O(12)
Worst Case: O(12)
Description: The space complexity is constant since we are only working with a fixed number of sections (12).
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