Appearance
Maximum Sum Triplet
Problem Description
Given an array A containing N integers.
You need to find the maximum sum of triplet ( Ai + Aj + Ak ) such that 0 <= i < j < k < N and Ai < Aj < Ak.
If no such triplet exist return 0.
Problem Constraints
3 <= N <= 10^5.
1 <= A[i] <= 10^8.
Input Format
First argument is an integer array A.
Output Format
Return a single integer denoting the maximum sum of triplet as described in the question.
Example Input
Input 1:
A = [2, 5, 3, 1, 4, 9]
Input 2:
A = [1, 2, 3]
Example Output
Output 1:
16
Output 2:
6
Example Explanation
Explanation 1:
All possible triplets are:-
2 3 4 => sum = 9
2 5 9 => sum = 16
2 3 9 => sum = 14
3 4 9 => sum = 16
1 4 9 => sum = 14
Maximum sum = 16
Explanation 2:
All possible triplets are:-
1 2 3 => sum = 6
Maximum sum = 6
Solution
swift
import Foundation
class Solution {
func insert(_ A: inout [Int], _ B:Int) -> Int {
var l = 0
var r = A.count-1
while l<=r {
var m = (l+r)/2
if A[m] == B {
A.insert(B, at:m)
while m >= 1 {
m -= 1
if A[m] < B{
return A[m]
}
}
return 0
} else if A[m] < B {
l = m + 1
} else {
r = m - 1
}
}
if r < 0 {
A.insert(B,at:0)
return 0
} else if l >= A.count {
A.append(B)
return A[A.count-2]
} else {
A.insert(B,at:l)
return A[r]
}
return 0
}
func solve(_ A: inout [Int]) -> Int {
let len = A.count
var ret = 0
var rights = [Int](repeating:0,count:len)
rights[len-1] = 0
for i in (0..<(len-1)).reversed() {
var r = max(A[i+1],rights[i+1])
if r > A[i]{
rights[i] = r
}
}
var lefts = [Int]()
lefts.append(A[0])
for i in 1..<(len-1) {
let l = insert(&lefts,A[i])
if l == 0 || rights[i] == 0 {
continue
}
ret = max(ret, l + A[i] + rights[i] )
}
return ret
}
}