How do you solve Sum of Total Strength of Wizards on LeetCode?

Sum of Total Strength of Wizards (LeetCode #2281) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each wizard's strength contributes to multiple subarrays, and we can calculate this contribution efficiently.

What is the brute force solution for Sum of Total Strength of Wizards?

The brute force approach for Sum of Total Strength of Wizards is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the total strength for every possible subarray by iterating through all pairs of indices. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Total Strength of Wizards?

Sum of Total Strength of Wizards uses the following concepts: Array, Stack, Monotonic Stack, Prefix Sum. The recommended patterns to study are: Monotonic Stack, Prefix Sum, Sliding Window.

What are the interview tips for Sum of Total Strength of Wizards?

Always think about how to reduce the number of calculations by using data structures like stacks or arrays. Practice explaining your thought process clearly, as communication is key in interviews. Be prepared to discuss the trade-offs of different approaches, especially time and space complexity.

What are common mistakes when solving Sum of Total Strength of Wizards?

Failing to consider edge cases, such as single-element subarrays. Not using modular arithmetic correctly, leading to overflow errors.

#2281

Sum of Total Strength of Wizards

Hard

ArrayStackMonotonic StackPrefix SumMonotonic StackPrefix SumSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the concept of prefix sums and a monotonic stack to efficiently calculate the contribution of each wizard to the total strength across all subarrays they are part of. This reduces the number of operations significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to store cumulative sums of strengths.
2Step 2: Use a monotonic stack to determine the next and previous smaller elements for each wizard's strength.
3Step 3: For each wizard, calculate their contribution to the total strength using the prefix sums and the indices from the monotonic stack.

solution.py21 lines

1def totalStrength(strength):
2    MOD = 10**9 + 7
3    n = len(strength)
4    prefix_sum = [0] * (n + 1)
5    for i in range(n):
6        prefix_sum[i + 1] = (prefix_sum[i] + strength[i]) % MOD
7    stack = []
8    total_sum = 0
9    for i in range(n):
10        while stack and strength[stack[-1]] > strength[i]:
11            j = stack.pop()
12            k = stack[-1] if stack else -1
13            total_sum += strength[j] * (prefix_sum[i] - prefix_sum[k + 1]) * (j - k) % MOD
14            total_sum %= MOD
15        stack.append(i)
16    while stack:
17        j = stack.pop()
18        k = stack[-1] if stack else -1
19        total_sum += strength[j] * (prefix_sum[n] - prefix_sum[k + 1]) * (j - k) % MOD
20        total_sum %= MOD
21    return total_sum

ℹ

Complexity note: The optimal solution runs in linear time due to the use of a monotonic stack, which allows us to efficiently find the contribution of each wizard without recalculating sums for overlapping subarrays.

1Each wizard's strength contributes to multiple subarrays, and we can calculate this contribution efficiently.
2Using prefix sums allows us to avoid recalculating sums for overlapping subarrays.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.