Valid Triangle Number LeetCode Solution

Difficulty Level Medium
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Problem Statement

Valid Triangle Number LeetCode Solution – 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.

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

Explanation

Once we sort the given  array, we need to find the right limit of the index k for a pair of indices (i, j) chosen to find the count of elements satisfying nums[i] + nums[j] > nums[k] for the triplet (nums[i], nums[j], nums[k]) to form a valid triangle.

We can find this right limit by simply traversing the index ‘s values starting from the index k=j+1 for a pair (i, j) chosen and stopping at the first value of k not satisfying the above inequality. Again, the count of elements nums[k] satisfying nums[i] + nums[j] > nums[k] for the pair of indices (i, j) chosen is given by k – j – 1 as discussed in the last approach.

Further, as discussed in the last approach, when we choose a higher value of index j for a particular i chosen, we need not start from the index j + 1. Instead, we can start off directly from the value of k where we left for the last index j. This helps to save redundant computations.

Valid Triangle Number LeetCode Solution

Code for Valid Triangle Number LeetCode Solution

Java Code:

public class Solution {
    public int triangleNumber(int[] nums) {
        int count = 0;
        Arrays.sort(nums);
        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[i] + nums[j] > nums[k])
                    k++;
                count += k - j - 1;
            }
        }
        return count;
    }
}

C++ Code:

class Solution {
public:
    int triangleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size(), ans = 0;
        for (int k = 2; k < n; ++k) {
            int i = 0, j = k - 1;
            while (i < j) {
                if (nums[i] + nums[j] > nums[k]) {
                    ans += j - i;
                    j -= 1;
                } else {
                    i += 1;
                }
            }
        }
        return ans;
    }
};

Complexity Analysis for Valid Triangle Number LeetCode Solution

Time Complexity

O(n^2)

Space Complexity

O(1)

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