# Towers Without a Valley - Google Top Interview Questions

### Problem Statement :

```You are given a list of integers nums. Consider a list of integers A such that A[i] ≤ nums[i]. Also, there are no j and k such that there exist j < i < k and A[j] > A[i] and A[i] < A[k].

Return the maximum possible sum of A.

Constraints

1 ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [10, 6, 8]

Output

22

Explanation

We can create A = [10, 6, 6].

Example 2

Input

nums = [4, 5, 1, 1, 5]

Output

12

Explanation

We can create A = [4, 5, 1, 1, 1].```

### Solution :

```                        ```Solution in C++ :

vector<int> calc(vector<int> a) {
int i, n = a.size(), sum = 0;
vector<int> ret(n, 0);
vector<pair<int, int>> st;
for (i = 0; i < n; i++) {
int count = 1;
sum += a[i];
while (st.size() && st.back().first >= a[i]) {
int num = st.back().first, freq = st.back().second;
count += freq;
sum -= (num - a[i]) * freq;
st.pop_back();
}
st.emplace_back(a[i], count);
ret[i] = sum;
}
return ret;
}

int solve(vector<int>& nums) {
vector<int> left = calc(nums);
reverse(nums.begin(), nums.end());
vector<int> right = calc(nums);
int ret = 0, i, n = nums.size();
for (i = 0; i < n; i++) ret = max(ret, left[i] + right[n - i - 1] - nums[n - 1 - i]);
return ret;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums) {
int n = nums.length;
int[] left = getIncreasingSum(nums);
int[] right = reverse(getIncreasingSum(reverse(nums)));
int max = Integer.MIN_VALUE;
for (int i = 0; i < n; i++) {
max = Math.max(max, left[i] + right[i] - nums[i]);
}

return max;
}

int[] reverse(int[] nums) {
int n = nums.length;
int[] res = new int[n];
for (int i = 0; i < n; i++) res[n - 1 - i] = nums[i];
return res;
}
int[] getIncreasingSum(int[] nums) {
int n = nums.length;
int[] res = new int[n];
Deque<Integer> dq = new ArrayDeque();
int total = 0;
for (int i = 0; i < nums.length; i++) {
while (!dq.isEmpty() && nums[dq.peekLast()] > nums[i]) {
int idx = dq.pollLast();
int count = !dq.isEmpty() ? idx - dq.peekLast() : idx + 1;
int diff = nums[idx] - nums[i];
total = total - (diff * count);
}
total += nums[i];
res[i] = total;
}
return res;
}
}```
```

```                        ```Solution in Python :

# Solves "half" of the problem:
# For each prefix `P` of `nums`, find the increasing sequence bounded by `P`
# with the maximum sum, and return that sum.
def increasing_prefix_sums(nums):
# Store the current best increasing sequence, in the form (value, number of occurrences)
stack = []
current_sum = 0  # always equal to sum(x * c for x in stack)

ans = []

for x in nums:
count = 1

# We need to include `x` in our sequence. Values to the left may need to be reduced.
while stack and stack[-1] > x:
# Anything greater than `x` gets clamped down to `x`.
# Since `stack` is increasing we only need to look at its tail.
# This is similar to the min-queue algorithm.
current_sum -= stack[-1] * stack[-1]
count += stack[-1]
stack.pop()

stack.append((x, count))
current_sum += x * count
ans.append(current_sum)

return ans

class Solution:
def solve(self, nums):
left = increasing_prefix_sums(nums)
right = reversed(increasing_prefix_sums(reversed(nums)))
# left[i] + right[i] overcounts by nums[i], so subtract that out
return max(l + r - x for l, r, x in zip(left, right, nums))```
```

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