**Sum of Three Numbers Trequel - Google Top Interview Questions**

### Problem Statement :

Given a list of positive integers nums, consider three indices i < j < k such that nums[i] ≤ nums[j] ≤ nums[k]. Return the maximum possible nums[i] + nums[j] + nums[k]. You can assume that a solution exists. Constraints 3 ≤ n ≤ 100,000 where n is the length of nums Example 1 Input nums = [9, 1, 5, 3, 4] Output 8 Explanation We pick [1, 3, 4] for a total sum of 8

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) {
int res = 0, sz = nums.size();
/*
* arr[i].first => some number in the number array
* arr[i].second => the corresponding index
* i.e. nums[ar[i].second] == ar[i].first
*/
vector<pair<int, int>> arr;
/*
* for each index i in the input(0 <= i < nums.length),
* larger[i] stores the number larger(or equal) than it on the right side.
* only numbers who have a larger/equal number to it on the right side can be the middle
* number a of valid triplet.
*/
unordered_map<int, int> larger;
// build the larger map for each index
for (long long i = sz - 1, mx = INT_MIN - 1LL; i >= 0; --i) {
if (mx >= nums[i]) larger[i] = mx;
mx = max(mx, (long long)nums[i]);
arr.push_back({nums[i], i});
}
sort(arr.begin(), arr.end());
stack<int> st;
/*
* We have our array arr sorted by value(i.e. arr[i].first), now for each i,
* we want to find a value such that it has smaller index in nums, it's value is smaller
* but is as large as possible
*/
for (int i = 0; i < sz; ++i) {
auto [num, idx] = arr[i];
if (larger.count(idx)) {
while (!st.empty() && arr[st.top()].second > idx) st.pop();
if (!st.empty()) res = max(res, nums[arr[st.top()].second] + num + larger[idx]);
st.push(i);
}
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] A) {
int res = Integer.MIN_VALUE;
int mx = Integer.MIN_VALUE;
int maxs[] = new int[A.length];
int B[] = new int[A.length];
Arrays.fill(B, Integer.MIN_VALUE);
for (int i = A.length - 1; i >= 0; i--) {
mx = Math.max(mx, A[i]);
maxs[i] = mx;
}
for (int i = A.length - 2; i >= 0; i--) {
if (maxs[i + 1] >= A[i]) {
B[i] = A[i] + maxs[i + 1];
}
}
TreeSet<Integer> tree = new TreeSet<>();
for (int i = 0; i < A.length; i++) {
if (B[i] != Integer.MIN_VALUE) {
Integer floor = tree.floor(A[i]);
if (floor != null) {
res = Math.max(res, floor + B[i]);
}
}
tree.add(A[i]);
}
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
max_sum = -(2 ** 31)
left_nums = SortedList(nums[:-1])
right_num = nums[-1]
for j in range(len(nums) - 2, 0, -1):
mid_num = nums[j]
left_nums.discard(mid_num)
if mid_num <= right_num:
i = left_nums.bisect_left(mid_num + 1) - 1
left_num = left_nums[i]
if left_num <= mid_num:
max_sum = max(max_sum, left_num + mid_num + right_num)
right_num = max(right_num, mid_num)
return max_sum
```

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