LeetCode 2012: Sum of Beauty in the Array Solution

Master LeetCode problem 2012 (Sum of Beauty in the Array), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2012. Sum of Beauty in the Array

Medium

Array

Problem Explanation

Explanation:

To solve this problem, we iterate over each element in the array from index 1 to nums.length - 2. For each element nums[i], we check if it satisfies the conditions for beauty being 2, 1, or 0 as described in the problem statement. We maintain two arrays leftMax and rightMin to keep track of the maximum element to the left of nums[i] and the minimum element to the right of nums[i] respectively. By using these arrays, we can efficiently determine the beauty of each element.

If nums[i] is greater than leftMax[i] and less than rightMin[i], then the beauty is 2.
If nums[i] is greater than nums[i-1] and less than nums[i+1], then the beauty is 1.
Otherwise, the beauty is 0.

Finally, we sum up all the beauty values to get the total sum of beauty in the array. :

Solution Code

class Solution {
    public int sumOfBeauties(int[] nums) {
        int n = nums.length;
        int[] leftMax = new int[n];
        int[] rightMin = new int[n];
        leftMax[0] = nums[0];
        rightMin[n-1] = nums[n-1];
        
        for (int i = 1; i < n; i++) {
            leftMax[i] = Math.max(leftMax[i-1], nums[i]);
        }
        
        for (int i = n-2; i >= 0; i--) {
            rightMin[i] = Math.min(rightMin[i+1], nums[i]);
        }
        
        int sum = 0;
        for (int i = 1; i < n-1; i++) {
            if (nums[i] > leftMax[i-1] && nums[i] < rightMin[i+1]) {
                sum += 2;
            } else if (nums[i] > nums[i-1] && nums[i] < nums[i+1]) {
                sum += 1;
            }
        }
        
        return sum;
    }
}

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