Find Total Power Amazon OA 2022

Find Total Power Amazon OA 2023 Solution

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Want to solve Find Total Power Amazon OA 2023?, if yes then this article is for you.

We have research and collected a bundle of questions which was asked in Amazon OA in the year of 2022 and 2023. Today, we are going to see another problem from amazon oa called Find Total Power which was asked in 2022, this questions might be still active in some of the Amazon OA so save all the solutions and lets see how can we solve this problem.

Find Total Power Amazon OA 2023 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.

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

Related: Google Online Assessment 2022 Questions List

Find Total Power Amazon OA 2022
Find Total Power Amazon OA 2023

SOLUTION

Program: Find Total Power Amazon OA Solution in Python

class Solution:
    def totalStrength(self, A: List[int]) -> int:
        N = len(A)
        Q = int(1e9)+7

        # monostack
        # ple[i] = previous lesser element, than the one at i
        st, ple = [], [-1]*N
        for i in range(N-1,-1,-1):
            while st and A[st[-1]] > A[i]:
                ple[st.pop()] = i
            st += [i]

        # preprocessing
        ps = [0] * (N+1)   # suffix sum
        ds = [0] * (N+1)   # suffix sum of suffix sum
        dpM = [0] * (N+1)  # dpM[i] = sum( suffixMin of A[0..i] )
        for z in range(N):
            i, j = ~z-1, ~z
            ps[i] = A[j] + ps[j]
            ds[i] = ps[i] + ds[j]
            k = ple[z]
            dpM[z] = A[z] * (z-k)
            if k != -1: dpM[z] += dpM[k]

        def query(L, R):  # -> sum( sum(A[k..R]) for k in [L..R] )
            if R < L: return 0
            diff = ds[L] - ds[R+1]
            chop = ps[R+1]
            size = R-L+1
            return diff - chop * size

        ans = 0
        dpC = [0] * (N+1)
        for i in range(N):
            j = ple[i]
            x = A[i] * query(j+1, i) % Q
            if j == -1: y = 0
            else:
                s = ps[j+1] - ps[i+1]
                y = (dpC[j] + dpM[j] * s) % Q
            dpC[i] = (x + y) % Q
            ans = (ans + dpC[i]) % Q

        return ans

Program: Find Total Power Amazon OA 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;
    }

Program: Find Total Power Amazon OA 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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