**Sum of Two Numbers Less Than Target - Amazon Top Interview Questions**

### Problem Statement :

Given a list of integers nums and an integer target, return the sum of the largest pair of numbers in nums whose sum is less than target. You can assume that there is a solution. Constraints 2 ≤ n ≤ 100,000 where n is the length of nums Example 1 Input nums = [5, 1, 2, 3] target = 4 Output 3 Explanation Sum of the largest pair of numbers less than 4 is 1 + 2 = 3.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ret = nums[0] + nums[1];
assert(ret < target);
int lhs = 0;
int rhs = nums.size() - 1;
while (lhs < rhs) {
int candSum = nums[lhs] + nums[rhs];
if (candSum >= target)
rhs--;
else {
ret = max(ret, candSum);
lhs++;
}
}
return ret;
}
```

` ````
Solution in Java :
import java.util.Arrays;
class Solution {
public int solve(int[] nums, int target) {
Arrays.sort(nums);
int i = 0, j = nums.length - 1, wanted = Integer.MIN_VALUE;
while (i < j) {
int sum = nums[i] + nums[j];
if (sum >= target) {
j--;
} else {
i++;
wanted = (sum > wanted) ? sum : wanted;
}
}
return wanted;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, target):
nums.sort()
p1 = 0
p2 = len(nums) - 1
m = -math.inf
while p1 < p2:
if nums[p1] + nums[p2] < target:
m = max(m, nums[p1] + nums[p2])
p1 += 1
else:
p2 -= 1
return m
```

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