Formatted question description: https://leetcode.ca/all/2281.html

2281. Sum of Total Strength of Wizards

• Difficulty: Hard.
• Related Topics: Array, Stack, Monotonic Stack, Prefix Sum.
• Similar Questions: Next Greater Element I, Sum of Subarray Minimums, Number of Visible People in a Queue, Sum of Subarray Ranges.

Problem

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.

A 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:
- [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.


Example 2:

Input: strength = [5,4,6]
Output: 213
Explanation: The following are all the contiguous groups of wizards:
- [5] from [5,4,6] has a total strength of min([5]) * sum([5]) = 5 * 5 = 25
- [4] from [5,4,6] has a total strength of min([4]) * sum([4]) = 4 * 4 = 16
- [6] from [5,4,6] has a total strength of min([6]) * sum([6]) = 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

Solution

• class Solution {
private static int mod = (int) 1e9 + 7;

public int totalStrength(int[] nums) {
int n = nums.length;
long[] forward = new long[n];
long[] backward = new long[n];
long[] prefix = new long[n + 1];
long[] suffix = new long[n + 1];
forward[0] = prefix[1] = nums[0];
backward[n - 1] = suffix[n - 1] = nums[n - 1];
for (int i = 1; i < n; ++i) {
forward[i] = nums[i] + forward[i - 1];
prefix[i + 1] = prefix[i] + forward[i];
}
for (int i = n - 2; 0 <= i; --i) {
backward[i] = nums[i] + backward[i + 1];
suffix[i] = suffix[i + 1] + backward[i];
}
long res = 0;
for (int i = 0; i < n; ++i) {
while (!dq.isEmpty() && nums[dq.peekLast()] >= nums[i]) {
int cur = dq.pollLast();
int prev = dq.isEmpty() ? -1 : dq.peekLast();
res =
(res
+ getSum(
nums, forward, prefix, backward, suffix,
prev, cur, i)
* nums[cur])
% mod;
}
}
while (!dq.isEmpty()) {
int cur = dq.pollLast();
int prev = dq.isEmpty() ? -1 : dq.peekLast();
res =
(res
+ getSum(nums, forward, prefix, backward, suffix, prev, cur, n)
* nums[cur])
% mod;
}
return (int) res;
}

private long getSum(
int[] nums,
long[] forward,
long[] prefix,
long[] backward,
long[] suffix,
int prev,
int cur,
int next) {
long sum = ((cur - prev) * (long) nums[cur] % mod) * (next - cur) % mod;
long preSum = getPresum(backward, suffix, prev + 1, cur - 1, next - cur);
long postSum = getPostsum(forward, prefix, cur + 1, next - 1, cur - prev);
return (sum + preSum + postSum) % mod;
}

private long getPresum(long[] backward, long[] suffix, int from, int to, int m) {
int n = backward.length;
long cnt = to - from + 1L;
return (suffix[from] - suffix[to + 1] - cnt * (to + 1 == n ? 0 : backward[to + 1]) % mod)
% mod
* m
% mod;
}

private long getPostsum(long[] forward, long[] prefix, int from, int to, int m) {
long cnt = to - from + 1L;
return (prefix[to + 1] - prefix[from] - cnt * (0 == from ? 0 : forward[from - 1]) % mod)
% mod
* m
% mod;
}
}

############

class Solution {
public int totalStrength(int[] strength) {
int n = strength.length;
int[] left = new int[n];
int[] right = new int[n];
Arrays.fill(left, -1);
Arrays.fill(right, n);
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
while (!stk.isEmpty() && strength[stk.peek()] >= strength[i]) {
stk.pop();
}
if (!stk.isEmpty()) {
left[i] = stk.peek();
}
stk.push(i);
}
stk.clear();
for (int i = n - 1; i >= 0; --i) {
while (!stk.isEmpty() && strength[stk.peek()] > strength[i]) {
stk.pop();
}
if (!stk.isEmpty()) {
right[i] = stk.peek();
}
stk.push(i);
}
int mod = (int) 1e9 + 7;
int[] s = new int[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = (s[i] + strength[i]) % mod;
}
int[] ss = new int[n + 2];
for (int i = 0; i < n + 1; ++i) {
ss[i + 1] = (ss[i] + s[i]) % mod;
}
long ans = 0;
for (int i = 0; i < n; ++i) {
int v = strength[i];
int l = left[i] + 1, r = right[i] - 1;
long a = (long) (i - l + 1) * (ss[r + 2] - ss[i + 1]);
long b = (long) (r - i + 1) * (ss[i + 1] - ss[l]);
ans = (ans + v * ((a - b) % mod)) % mod;
}
return (int) (ans + mod) % mod;
}
}

• 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

############

# 2281. Sum of Total Strength of Wizards
# https://leetcode.com/problems/sum-of-total-strength-of-wizards

class Solution:
def totalStrength(self, A: List[int]) -> int:
n = len(A)

right = [n] * n
stack = []

for i in range(n):
while stack and A[i] < A[stack[-1]]:
right[stack.pop()] = i

stack.append(i)

left = [-1] * n
stack = []

for i in range(n - 1, -1, -1):
while stack and A[i] <= A[stack[-1]]:
left[stack.pop()] = i

stack.append(i)

res = 0
M = 10 ** 9 + 7
acc = list(accumulate(accumulate(A, initial = 0)))

for i, x in enumerate(A):
l, r = left[i], right[i]
leftSum = acc[i] - acc[max(0, l)]
rightSum = acc[r] - acc[i]
ln, rn = i - l, r - i
res += A[i] * (rightSum * ln - leftSum * rn)
res %= M

return res % M


• class Solution {
public:
int totalStrength(vector<int>& strength) {
int n = strength.size();
vector<int> left(n, -1);
vector<int> right(n, n);
stack<int> stk;
for (int i = 0; i < n; ++i) {
while (!stk.empty() && strength[stk.top()] >= strength[i]) stk.pop();
if (!stk.empty()) left[i] = stk.top();
stk.push(i);
}
stk = stack<int>();
for (int i = n - 1; i >= 0; --i) {
while (!stk.empty() && strength[stk.top()] > strength[i]) stk.pop();
if (!stk.empty()) right[i] = stk.top();
stk.push(i);
}
int mod = 1e9 + 7;
vector<int> s(n + 1);
for (int i = 0; i < n; ++i) s[i + 1] = (s[i] + strength[i]) % mod;
vector<int> ss(n + 2);
for (int i = 0; i < n + 1; ++i) ss[i + 1] = (ss[i] + s[i]) % mod;
int ans = 0;
for (int i = 0; i < n; ++i) {
int v = strength[i];
int l = left[i] + 1, r = right[i] - 1;
long a = (long) (i - l + 1) * (ss[r + 2] - ss[i + 1]);
long b = (long) (r - i + 1) * (ss[i + 1] - ss[l]);
ans = (ans + v * ((a - b) % mod)) % mod;
}
return (int) (ans + mod) % mod;
}
};

• func totalStrength(strength []int) int {
n := len(strength)
left := make([]int, n)
right := make([]int, n)
for i := range left {
left[i] = -1
right[i] = n
}
stk := []int{}
for i, v := range strength {
for len(stk) > 0 && strength[stk[len(stk)-1]] >= v {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
left[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
stk = []int{}
for i := n - 1; i >= 0; i-- {
for len(stk) > 0 && strength[stk[len(stk)-1]] > strength[i] {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
right[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
mod := int(1e9) + 7
s := make([]int, n+1)
for i, v := range strength {
s[i+1] = (s[i] + v) % mod
}
ss := make([]int, n+2)
for i, v := range s {
ss[i+1] = (ss[i] + v) % mod
}
ans := 0
for i, v := range 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 + v*((a-b)%mod)) % mod
}
return (ans + mod) % mod
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).