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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:
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The strength of the weakest wizard in the group.
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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
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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; Deque<Integer> dq = new LinkedList<>(); 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; } dq.add(i); } 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).