Russian Doll Envelopes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given a 2D array of integers envelopes where envelopes[i] = [wi, hi] represents the width and the height of an envelope.
One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope’s width and height.
Return the maximum number of envelopes you can Russian doll (i.e., put one inside the other).
Note: You cannot rotate an envelope.
Example 1:
Input: envelopes = [[5,4],[6,4],[6,7],[2,3]]
Output: 3
Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).
Example 2:
Input: envelopes = [[1,1],[1,1],[1,1]] Output: 1
Constraints:
1 <= envelopes.length <= 105envelopes[i].length == 21 <= wi, hi <= 105
C++ Russian Doll Envelopes LeetCode Solution
struct Solution {
int maxEnvelopes(vector<pair<int, int>>& envelopes) {
if (envelopes.empty()) return 0;
sort(envelopes.begin(), envelopes.end());
vector<int> dp(envelopes.size(), 1);
for (int i = 0; i < envelopes.size(); ++i)
for (int j = 0; j < i; ++j)
if (envelopes[j].first < envelopes[i].first && envelopes[j].second < envelopes[i].second)
dp[i] = max(dp[i] , dp[j] + 1);
return *max_element(dp.begin(), dp.end());
}
};
Java Russian Doll Envelopes LeetCode Solution
public int maxEnvelopes(int[][] envelopes) {
if(envelopes == null || envelopes.length == 0
|| envelopes[0] == null || envelopes[0].length != 2)
return 0;
Arrays.sort(envelopes, new Comparator<int[]>(){
public int compare(int[] arr1, int[] arr2){
if(arr1[0] == arr2[0])
return arr2[1] - arr1[1];
else
return arr1[0] - arr2[0];
}
});
int dp[] = new int[envelopes.length];
int len = 0;
for(int[] envelope : envelopes){
int index = Arrays.binarySearch(dp, 0, len, envelope[1]);
if(index < 0)
index = -(index + 1);
dp[index] = envelope[1];
if(index == len)
len++;
}
return len;
}
Python 3 Russian Doll Envelopes LeetCode Solution
class Solution(object):
def maxEnvelopes(self, envs):
def liss(envs):
def lmip(envs, tails, k):
b, e = 0, len(tails) - 1
while b <= e:
m = (b + e) >> 1
if envs[tails[m]][1] >= k[1]:
e = m - 1
else:
b = m + 1
return b
tails = []
for i, env in enumerate(envs):
idx = lmip(envs, tails, env)
if idx >= len(tails):
tails.append(i)
else:
tails[idx] = i
return len(tails)
def f(x, y):
return -1 if (x[0] < y[0] or x[0] == y[0] and x[1] > y[1]) else 1
envs.sort(cmp=f)
return liss(envs)
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