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611.valid-triangle-number.cpp
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68 lines (65 loc) · 1.57 KB
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// Tag: Array, Two Pointers, Binary Search, Greedy, Sorting
// Time: O(N^2)
// Space: O(1)
// Ref: -
// Note: -
// Video: https://youtu.be/bk7wSRuKgco
// Given an integer array nums, return 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: nums = [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
//
// Example 2:
//
// Input: nums = [4,2,3,4]
// Output: 4
//
//
// Constraints:
//
// 1 <= nums.length <= 1000
// 0 <= nums[i] <= 1000
//
//
class Solution {
public:
int triangleNumber(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
int res = 0;
for (int k = 2; k < n; k++) {
int l = 0;
int r = k - 1;
while (l < r) {
if (nums[l] + nums[r] > nums[k]) {
res += r - l;
r -= 1;
} else {
l += 1;
}
}
}
return res;
}
};
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
int res = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
int k = lower_bound(nums.begin(), nums.end(), nums[i] + nums[j]) - nums.begin();
res += max(0, k - j - 1);
}
}
return res;
}
};