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# Sum of Total Strength of Wizards LeetCode Solution

## Problem – Sum of Total Strength of Wizards leetCode Solution

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. For a contiguous group of wizards (i.e. the wizards’ strengths form a subarray of `strength`), the total strength is defined as the product of the following two values:

• The strength of the weakest wizard in the group.
• The total of all the individual strengths of the wizards in the group.

Return the sum of the total strengths of all contiguous groups of wizards. Since the answer may be very large, return it modulo `109 + 7`.

subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

``````Input: strength = [1,3,1,2]
Output: 44
Explanation: The following are all the contiguous groups of wizards:
-  from [1,3,1,2] has a total strength of min() * sum() = 1 * 1 = 1
-  from [1,3,1,2] has a total strength of min() * sum() = 3 * 3 = 9
-  from [1,3,1,2] has a total strength of min() * sum() = 1 * 1 = 1
-  from [1,3,1,2] has a total strength of min() * sum() = 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.
``````

Example 2:

``````Input: strength = [5,4,6]
Output: 213
Explanation: The following are all the contiguous groups of wizards:
-  from [5,4,6] has a total strength of min() * sum() = 5 * 5 = 25
-  from [5,4,6] has a total strength of min() * sum() = 4 * 4 = 16
-  from [5,4,6] has a total strength of min() * sum() = 6 * 6 = 36
- [5,4] from [5,4,6] has a total strength of min([5,4]) * sum([5,4]) = 4 * 9 = 36
- [4,6] from [5,4,6] has a total strength of min([4,6]) * sum([4,6]) = 4 * 10 = 40
- [5,4,6] from [5,4,6] has a total strength of min([5,4,6]) * sum([5,4,6]) = 4 * 15 = 60
The sum of all the total strengths is 25 + 16 + 36 + 36 + 40 + 60 = 213.
``````

Constraints:

• `1 <= strength.length <= 105`
• `1 <= strength[i] <= 109`

### Sum of Total Strength of Wizards LeetCode Solution in Python

``````    def totalStrength(self, A):
mod = 10 ** 9 + 7
n = len(A)

# next small on the right
right = [n] * n
stack = []
for i in range(n):
while stack and A[stack[-1]] > A[i]:
right[stack.pop()] = i
stack.append(i)

# next small on the left
left = [-1] * n
stack = []
for i in range(n-1, -1, -1):
while stack and A[stack[-1]] >= A[i]:
left[stack.pop()] = i
stack.append(i)

# for each A[i] as minimum, calculate sum
res = 0
acc = list(accumulate(accumulate(A), initial = 0))
for i in range(n):
l, r = left[i], right[i]
lacc = acc[i] - acc[max(l, 0)]
racc = acc[r] - acc[i]
ln, rn = i - l, r - i
res += A[i] * (racc * ln - lacc * rn)
return res
``````

### Sum of Total Strength of Wizards LeetCode Solution in Java

``````    public int totalStrength(int[] A) {
int res = 0, ac = 0, mod = (int)1e9 + 7, n = A.length;
Stack<Integer> stack = new Stack<>();
int[] acc = new int[n + 2];
for (int r = 0; r <= n; ++r) {
int a = r < n ? A[r] : 0;
ac = (ac + a) % mod;
acc[r + 1] = (ac + acc[r]) % mod;
while (!stack.isEmpty() && A[stack.peek()] > a) {
int i = stack.pop();
int l = stack.isEmpty() ? -1 : stack.peek();
long lacc = l < 0 ? acc[i] : acc[i] - acc[l], racc = acc[r] - acc[i];
int ln = i - l, rn = r - i;
res = (int)(res + (racc * ln - lacc * rn) % mod * A[i] % mod) % mod;
}
stack.push(r);
}
return (res + mod) % mod;
}
``````

### Sum of Total Strength of Wizards LeetCode Solution in C++

``````    int totalStrength(vector<int>& A) {
int res = 0, ac = 0, mod = 1e9 + 7, n = A.size();
vector<int> stack = {}, acc(n + 2);
for (int r = 0; r <= n; ++r) {
int a = r < n ? A[r] : 0;
ac = (ac + a) % mod;
acc[r + 1] = (ac + acc[r]) % mod;
while (!stack.empty() && A[stack.back()] > a) {
int i = stack.back(); stack.pop_back();
int l = stack.empty() ? -1 : stack.back();
long lacc = l < 0 ? acc[i] : acc[i] - acc[l], racc = acc[r] - acc[i];
int ln = i - l, rn = r - i;
res = (res + (racc * ln - lacc * rn) % mod * A[i] % mod) % mod;
}
stack.push_back(r);
}
return (res + mod) % mod;
}
``````
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