Leetcode #2281: Sum of Total Strength of Wizards
In this guide, we solve Leetcode #2281 Sum of Total Strength of Wizards in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
As the ruler of a kingdom, you have an army of wizards at your command. You are given a 0-indexed integer array strength, where strength[i] denotes the strength of the ith wizard.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Array, Prefix Sum, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: strength = [1,3,1,2]
Output: 44
Explanation: The following are all the contiguous groups of wizards:
- [1] from [1,3,1,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
- [3] from [1,3,1,2] has a total strength of min([3]) * sum([3]) = 3 * 3 = 9
- [1] from [1,3,1,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
- [2] from [1,3,1,2] has a total strength of min([2]) * sum([2]) = 2 * 2 = 4
- [1,3] from [1,3,1,2] has a total strength of min([1,3]) * sum([1,3]) = 1 * 4 = 4
- [3,1] from [1,3,1,2] has a total strength of min([3,1]) * sum([3,1]) = 1 * 4 = 4
- [1,2] from [1,3,1,2] has a total strength of min([1,2]) * sum([1,2]) = 1 * 3 = 3
- [1,3,1] from [1,3,1,2] has a total strength of min([1,3,1]) * sum([1,3,1]) = 1 * 5 = 5
- [3,1,2] from [1,3,1,2] has a total strength of min([3,1,2]) * sum([3,1,2]) = 1 * 6 = 6
- [1,3,1,2] from [1,3,1,2] has a total strength of min([1,3,1,2]) * sum([1,3,1,2]) = 1 * 7 = 7
The sum of all the total strengths is 1 + 9 + 1 + 4 + 4 + 4 + 3 + 5 + 6 + 7 = 44.
Python Solution
class Solution:
def totalStrength(self, strength: List[int]) -> int:
n = len(strength)
left = [-1] * n
right = [n] * n
stk = []
for i, v in enumerate(strength):
while stk and strength[stk[-1]] >= v:
stk.pop()
if stk:
left[i] = stk[-1]
stk.append(i)
stk = []
for i in range(n - 1, -1, -1):
while stk and strength[stk[-1]] > strength[i]:
stk.pop()
if stk:
right[i] = stk[-1]
stk.append(i)
ss = list(accumulate(list(accumulate(strength, initial=0)), initial=0))
mod = int(1e9) + 7
ans = 0
for i, v in enumerate(strength):
l, r = left[i] + 1, right[i] - 1
a = (ss[r + 2] - ss[i + 1]) * (i - l + 1)
b = (ss[i + 1] - ss[l]) * (r - i + 1)
ans = (ans + (a - b) * v) % mod
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.