[LeetCode] 2212. Maximum Points in an Archery Competition explained
![title image](/public/images/leetcode-teaser.png)
Problem
2212. Maximum Points in an Archery Competition
Approach
Note that we have fairly short range of points { 0, 1, ... 11 }
.
Total number of subsets is \(2^{12} = 4096\), and each of them can be Bob’s choice as long as he has enough arrows.
Brute force all the cases using bitmask
, and find the maximum total point of Bob’s satisfying certain condition:
int maxPoint= 0, maxMask;
for (int mask=0; mask < (1<<12); ++mask) {
if (totalPoint(mask) > maxPoint &&
arrowsNeeded(mask, aliceArrows) <= numArrows) {
maxPoint= totalPoint(mask);
maxMask= mask;
}
}
Helper functions:
private:
int totalPoint(int mask) {
int point= 0;
for (int i=0; i < 12; ++i) {
if (mask & (1<<i)) point += i;
}
return point;
}
int arrowsNeeded(int mask, vi &aliceArrows) {
int arrows= 0;
for (int i=0; i < 12; ++i) {
if (mask & (1<<i)) {
arrows += aliceArrows[i]+1;
}
}
return arrows;
}
Now that you have the maxMask
, construct the answer array:
vi ret(12,0);
int cnt= numArrows;
for (int i=1; i < 12; ++i) {
if (maxMask & (1<<i)) {
ret[i]= aliceArrows[i]+1;
cnt -= ret[i];
}
}
ret[0]= cnt;
Code
typedef vector<int> vi;
class Solution {
private:
int totalPoint(int mask) {
int point= 0;
for (int i=0; i < 12; ++i) {
if (mask & (1<<i)) point += i;
}
return point;
}
int arrowsNeeded(int mask, vi &aliceArrows) {
int arrows= 0;
for (int i=0; i < 12; ++i) {
if (mask & (1<<i)) {
arrows += aliceArrows[i]+1;
}
}
return arrows;
}
public:
vi maximumBobPoints(int numArrows, vi &aliceArrows) {
int maxPoint= 0, maxMask;
for (int mask=0; mask < (1<<12); ++mask) {
if (totalPoint(mask) > maxPoint &&
arrowsNeeded(mask, aliceArrows) <= numArrows) {
maxPoint= totalPoint(mask);
maxMask= mask;
}
}
vi ret(12,0);
int cnt= numArrows;
for (int i=1; i < 12; ++i) {
if (maxMask & (1<<i)) {
ret[i]= aliceArrows[i]+1;
cnt -= ret[i];
}
}
ret[0]= cnt;
return ret;
}
};
Complexity
- Time: \(O(1)\)
- Space: \(O(1)\)
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