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Last Stone Weight II in C++
Suppose we have a collection of rocks, now each rock has a positive integer weight. In each turn, we choose any two rocks and smash them together. If the stones have weights x and y with x <= y. The result of this smash will be −
If x = y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
Finally, there is at most 1 stone left. We have to find the smallest possible weight of this stone (the weight is 0 if there are no stones left.)
So for example, if the input is like [2,7,4,1,8,1], then the output will be 1. This is because if we smash (2,4), then the new array will be [2,7,1,8,1], them smash (7,8), the new array will be [2,1,1,1], then smash (2,1), the array will be [1,1,1], after that smash (1,1), so only 1 will be there.
To solve this, we will follow these steps −
n := size of the stones array, set total := 0
for i in range 0 to n – 1
total := total + stones[i]
req := total / 2
make an array dp of size req + 1, and fill this with false values
dp[0] := true, reach := 0
for i in range 0 to n – 1
for j := req, when j – stones[i] >= 0, decrease j by 1
dp[j] := false when (dp[j] and dp[j – stones[i]]) both are false, otherwise true
if dp[j] is true, then reach := max of reach and j
return total – (2 * reach)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int lastStoneWeightII(vector<int>& stones) {
int n = stones.size();
int total = 0;
for(int i = 0; i < n; i++){
total += stones[i];
}
int req = total / 2;
vector <bool> dp(req + 1, false);
dp[0] = true;
int reach = 0;
for(int i = 0; i < n; i++){
for(int j = req; j - stones[i] >= 0; j--){
dp[j] = dp[j] || dp[j - stones[i]];
if(dp[j]) reach = max(reach, j);
}
}
return total - (2 * reach);
}
};
main(){
vector<int> v = {2,7,4,1,8,1};
Solution ob;
cout << (ob.lastStoneWeightII(v));
}Input
[2,7,4,1,8,1]
Output
1
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