**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];
dq.addLast(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][0] > 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][0] * stack[-1][1]
count += stack[-1][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))
```

## View More Similar Problems

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

View Solution →## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

View Solution →