Algorithm
Problem Name: 611. Valid Triangle Number
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 <= 10000 <= nums[i] <= 1000
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int count = 0, n = nums.size();
for(int i = 0; i < n; i++)
for(int j = i + 1; j < n; j++)
for(int k = j + 1; k < n; k++)
if(nums[i] + nums[j] > nums[k]) count++;
return count;
}
};
class Solution {
public:
int triangleNumber(vector<int>& nums) {
sort(nums.begin(), nums.end());
int count = 0, n = nums.size();
for(int i = n - 1; i > 1; --i) {
int l = 0, r = i - 1;
while (l < r> {
if (nums[l] + nums[r] > nums[i]) {
count += r - l;
--r;
} else {
++l;
}
}
}
return count;
}
};
Input
nums = [2,2,3,4]
Output
3
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int count = 0;
for (int i=0;i < nums.length-2;i++) {
int k = i+2;
for (int j=i+1;j < nums.length-1 && nums[i] != 0; j++) {
while(k < nums.length && nums[k] < nums[i] + nums[j])
k++;
count += k-j-1;
}
}
return count;
}
}
Input
nums = [2,2,3,4]
Output
3
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const triangleNumber = function(nums) {
nums.sort((a, b) => a - b);
let count = 0;
let n = nums.length;
for (let i = n - 1; i >= 2; i--) {
let lo = 0;
let mid = i - 1;
while (lo < mid) {
if (nums[lo] + nums[mid] > nums[i]) {
count += mid - lo;
mid -= 1;
} else {
lo += 1;
}
}
}
return count;
};
console.log(triangleNumber([2, 2, 3, 4]));
Input
nums = [4,2,3,4]
Output
4