LeetCode #2012 — MEDIUM

Sum of Beauty in the Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2) the beauty of nums[i] equals:

2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1.
1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied.
0, if none of the previous conditions holds.

Return the sum of beauty of all nums[i] where 1 <= i <= nums.length - 2.

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: For each index i in the range 1 <= i <= 1:
- The beauty of nums[1] equals 2.

Example 2:

Input: nums = [2,4,6,4]
Output: 1
Explanation: For each index i in the range 1 <= i <= 2:
- The beauty of nums[1] equals 1.
- The beauty of nums[2] equals 0.

Example 3:

Input: nums = [3,2,1]
Output: 0
Explanation: For each index i in the range 1 <= i <= 1:
- The beauty of nums[1] equals 0.

Constraints:

3 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2) the beauty of nums[i] equals: 2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1. 1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied. 0, if none of the previous conditions holds. Return the sum of beauty of all nums[i] where 1 <= i <= nums.length - 2.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,2,3]

Example 2

[2,4,6,4]

Example 3

[3,2,1]

Core Insight

What unlocks the optimal approach

Use suffix/prefix arrays.
prefix[i] records the maximum value in range (0, i - 1) inclusive.
suffix[i] records the minimum value in range (i + 1, n - 1) inclusive.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2012: Sum of Beauty in the Array
class Solution {
    public int sumOfBeauties(int[] nums) {
        int n = nums.length;
        int[] right = new int[n];
        right[n - 1] = nums[n - 1];
        for (int i = n - 2; i > 0; --i) {
            right[i] = Math.min(right[i + 1], nums[i]);
        }
        int ans = 0;
        int l = nums[0];
        for (int i = 1; i < n - 1; ++i) {
            int r = right[i + 1];
            if (l < nums[i] && nums[i] < r) {
                ans += 2;
            } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
                ans += 1;
            }
            l = Math.max(l, nums[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2012: Sum of Beauty in the Array
func sumOfBeauties(nums []int) (ans int) {
	n := len(nums)
	right := make([]int, n)
	right[n-1] = nums[n-1]
	for i := n - 2; i > 0; i-- {
		right[i] = min(right[i+1], nums[i])
	}
	for i, l := 1, nums[0]; i < n-1; i++ {
		r := right[i+1]
		if l < nums[i] && nums[i] < r {
			ans += 2
		} else if nums[i-1] < nums[i] && nums[i] < nums[i+1] {
			ans++
		}
		l = max(l, nums[i])
	}
	return
}

# Accepted solution for LeetCode #2012: Sum of Beauty in the Array
class Solution:
    def sumOfBeauties(self, nums: List[int]) -> int:
        n = len(nums)
        right = [nums[-1]] * n
        for i in range(n - 2, -1, -1):
            right[i] = min(right[i + 1], nums[i])
        ans = 0
        l = nums[0]
        for i in range(1, n - 1):
            r = right[i + 1]
            if l < nums[i] < r:
                ans += 2
            elif nums[i - 1] < nums[i] < nums[i + 1]:
                ans += 1
            l = max(l, nums[i])
        return ans

// Accepted solution for LeetCode #2012: Sum of Beauty in the Array
impl Solution {
    pub fn sum_of_beauties(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut right: Vec<i32> = vec![0; n];
        right[n - 1] = nums[n - 1];
        for i in (1..n - 1).rev() {
            right[i] = right[i + 1].min(nums[i]);
        }
        let mut ans = 0;
        let mut l = nums[0];
        for i in 1..n - 1 {
            let r = right[i + 1];
            if l < nums[i] && nums[i] < r {
                ans += 2;
            } else if nums[i - 1] < nums[i] && nums[i] < nums[i + 1] {
                ans += 1;
            }
            l = l.max(nums[i]);
        }
        ans
    }
}

// Accepted solution for LeetCode #2012: Sum of Beauty in the Array
function sumOfBeauties(nums: number[]): number {
    const n = nums.length;
    const right: number[] = Array(n).fill(nums[n - 1]);
    for (let i = n - 2; i; --i) {
        right[i] = Math.min(right[i + 1], nums[i]);
    }
    let ans = 0;
    for (let i = 1, l = nums[0]; i < n - 1; ++i) {
        const r = right[i + 1];
        if (l < nums[i] && nums[i] < r) {
            ans += 2;
        } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
            ans += 1;
        }
        l = Math.max(l, nums[i]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.