Class Solution {
Public Boolean Flipequiv(Treenode Root1, Treenode Root2) {
Return Dfs(Root1, Root2);
}
Private Boolean Dfs(Treenode Root1, Treenode Root2) {
If (Root1 == Root2 || (Root1 == Null && Root2 == Null)) {
Return True;
}
If (Root1 == Null Root2 == Null Root1.Val != Root2.Val) {
Return False;
}
Return (Dfs(Root1.Left, Root2.Left) && Dfs(Root1.Right, Root2.Right))
|| (Dfs(Root1.Left, Root2.Right) && Dfs(Root1.Right, Root2.Left));
}
}
Public Boolean Flipequiv(Treenode Root1, Treenode Root2) {
Return Dfs(Root1, Root2);
}
Private Boolean Dfs(Treenode Root1, Treenode Root2) {
If (Root1 == Root2 || (Root1 == Null && Root2 == Null)) {
Return True;
}
If (Root1 == Null
Return False;
}
Return (Dfs(Root1.Left, Root2.Left) && Dfs(Root1.Right, Root2.Right))
|| (Dfs(Root1.Left, Root2.Right) && Dfs(Root1.Right, Root2.Left));
}
}
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334. Increasing Triplet Subsequence
Medium
Topics
Companies
Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
Example 1:
Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.
Example 3:
Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.class Solution {
public boolean increasingTriplet(int[] nums) {
int first = Integer.MAX_VALUE;
int second = Integer.MAX_VALUE;
for (int num : nums) {
if (num <= first) {
first = num;
} else if (num <= second) {
second = num;
} else {
return true;
}
}
return false;
}
}