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Bad Neighbors TopCoder
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/* Update - Thanks to some folks who pointed out a flaw in my code. | |
* I have updated it. Hope it is correct now :-) | |
* Old version can be found here - | |
https://gist.github.com/krisys/4089748/262cbc10d9b9f1cb5df771e14a1e143a86d2ecc6/ | |
*/ | |
#include <iostream> | |
#include <vector> | |
#include <algorithm> | |
using namespace std; | |
class BadNeighbors{ | |
public: | |
int maxDonations(vector <int> amount); | |
}; | |
int BadNeighbors::maxDonations(vector <int> amount){ | |
int N = amount.size(); | |
if (N < 3){ | |
return *max(amount.begin(), amount.end()); | |
} | |
/* contains the best value till index i */ | |
vector <int> dp(50); | |
/* Flag to denote if 0th element was included to get this value */ | |
vector <bool> zero_included(50, false) | |
/* Flag to denote if ith element was included to get dp[i] */ | |
vector <bool> i_included(50, false); | |
// Initialize first two elements | |
dp[0] = amount[0]; | |
i_included[0] = zero_included[0] = true; | |
if (amount[1] > dp[0]){ | |
dp[1] = amount[1]; | |
i_included[1] = true; | |
} else { | |
dp[1] = dp[0]; | |
i_included[1] = false; | |
zero_included[1] = zero_included[0]; | |
} | |
for(int i = 2; i < N; i++){ | |
if(dp[i-2] + amount[i] > dp[i-1]){ | |
dp[i] = dp[i-2] + amount[i]; | |
i_included[i] = true; | |
zero_included[i] = zero_included[i-2]; | |
} else { | |
dp[i] = dp[i-1]; | |
i_included[i] = false; | |
zero_included[i] = zero_included[i-1]; | |
} | |
} | |
if(i_included[N-1] && zero_included[N-1]){ | |
return max( | |
max( | |
dp[N-2], | |
dp[N-1] - amount[N-1] | |
), | |
dp[N-1] - amount[0] | |
); | |
} else { | |
return dp[N-1]; | |
} | |
} | |
int donations[] = { 10, 3, 2, 5, 7, 8 }; | |
/*int donations[] = { 11, 15 };*/ | |
/*int donations[] = { 7, 7, 7, 7, 7, 7, 7 };*/ | |
/*int donations[] = { 1, 2, 3, 4, 5, 1, 2, 3, 4, 5 };*/ | |
/*int donations[] = { 94, 40, 49, 65, 21, 21, 106, 80, 92, 81, 679, 4, 61, | |
6, 237, 12, 72, 74, 29, 95, 265, 35, 47, 1, 61, 397, | |
52, 72, 37, 51, 1, 81, 45, 435, 7, 36, 57, 86, 81, 72 };*/ | |
/*int donations[] = {10, 3, 2, 5, 1};*/ | |
/*int donations[] = {590, 260, 60, 331, 929, 661, 675, 181, 593, | |
69, 253, 675, 410, 328, 734, 560, 547, 520, 408, 108, 346, 14, | |
982, 808, 944, 896, 558, 691, 941, 341, 306, 471, 750, 592, | |
279, 205, 485, 483, 895, 415}; */ | |
/*int donations[] = {442, 851, 111, 479, 261, 770, 194, 619, 864, 263, | |
323, 15, 110, 37, 926, 399, 916, 824, 811, 988, 290, 135, 633, 766, | |
293, 554, 193, 943, 480, 239, 391, 865, 778, 962, 383, 792, 42, | |
836, 646, 727};*/ | |
/*int donations[] = {90, 1, 2, 89, 3, 4, 88, 5, 6, 87, 5, 4, 86, 3, 2, 85}; */ | |
/*int donations[] = {307, 322, 35, 140, 667, 1, 884, 44, 546, 239, | |
548, 881, 222, 459, 55, 158, 642, 280, 237, 561, 925, 539, 38, | |
699, 107, 116, 449, 418, 720, 487, 983, 743}; */ | |
/*int donations[] = {917, 19, 978, 687, 346, 245, 677, 708, 565, 940, | |
211, 127, 993, 768, 469, 279, 460, 335, 734, 130, 691, 783, 182, | |
391, 827, 295, 534, 263, 193, 498, 327, 120, 690 | |
};*/ | |
int main(int argc, char const *argv[]){ | |
BadNeighbors BN; | |
int N = sizeof(donations)/ sizeof(donations[0]); | |
cout << BN.maxDonations(vector <int> (donations, donations + N)) << endl; | |
} |
if(i_included[N-1] && zero_included[N-1]){
return max(
max(
dp[N-2],
dp[N-1] - amount[N-1]
),
dp[N-1] - amount[0]
);
} else {
return dp[N-1];
}
what is meant by this segment of code....
if you can't use the last number because the first number is also used, then return max of
- optimal solution for second to last number
- optimal solution for last number, excluding the last number
- optimal solution for last number, excluding the first number
If the first and last numbers aren't used, just return the optimal solution for the last number
in this :int donations[] = { 10, 3, 2, 5, 7, 8 ,1};
result is 20 (10+2+7+1),but ,is the answer should be 24 (10+5+8+1) ??
Shouldn't the answer be 23 ???? (10 + 5 + 8)
If I am here hopefully more will come, so dropping a trick I learned some time ago regarding the questions that involve circle.
#include <bits/stdc++.h>
using namespace std;
class BadNeighbors{
public:
int max_donations(vector<int> donations, int start, int end) {
int n = donations.size();
vector<int> dp(n + 1, 0);
dp[start] = donations[start];
dp[start + 1] = donations[start + 1];
for(int i = start + 2; i < end; i++) {
dp[i] = max(donations[i] + dp[i - 2], dp[i - 1]);
}
return max(dp[end - 1], dp[end - 2]);
}
int maxDonations(vector<int> donations) {
int n = donations.size();
if (n == 1) return donations[0];
if (n == 2) return max(donations[0], donations[1]);
return max(max_donations(donations, 1, n), max_donations(donations, 0, n - 1));
}
};
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you don't need O(n) space to resolve this question.