Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
Example 1:
Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
Note:
- The length of the given array won't exceed 1000.
- The integers in the given array are in the range of [0, 1000].
The naive solution is of O(N^3) time complexity, that is, for each triplet, detect if it can form a triangle. This solution will get TLE.
To optimize it, I first sort nums in ascending order. And for each doublet a and b, use binary search to find the count of numbers greater than a + b and less than a - b (a >= b).
// OJ: https://leetcode.com/problems/valid-triangle-number
// Author: github.com/lzl124631x
// Time: O(N^2logN)
// Space: O(1)
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cnt = 0, N = nums.size();
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
int lb = nums[j] - nums[i], rb = nums[i] + nums[j];
int L = j + 1, R = N - 1, left = 0;
while (L <= R) {
int M = (L + R) / 2;
if (nums[M] > lb) R = M - 1;
else L = M + 1;
}
left = L;
L = j + 1, R = N - 1;
while (L <= R) {
int M = (L + R) / 2;
if (nums[M] >= rb) R = M - 1;
else L = M + 1;
}
if (R >= left) cnt += R - left + 1;
}
}
return cnt;
}
};Same as solution 1, just uses built-in functions lower_bound and upper_bound.
// OJ: https://leetcode.com/problems/valid-triangle-number
// Author: github.com/lzl124631x
// Time: O(N^2logN)
// Space: O(1)
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cnt = 0, N = nums.size();
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j < N; ++j) {
auto left = upper_bound(nums.begin() + j + 1, nums.end(), nums[j] - nums[i]);
auto right = lower_bound(nums.begin() + j + 1, nums.end(), nums[i] + nums[j]);
if (right > left) cnt += right - left;
}
}
return cnt;
}
};