# Swift O(n) time solution

• Turn the number to a list of digits. Traverse backwards, for each digit d, if d is smaller than the maximum M of all digits behind it, then d may be swapped with M to obtain a larger number.

The idea is that the last (d, M) pair found is the pair to swap.
It may be explained as a greedy approach.

Update indices of (d, M) when such pair is found. Check that index of d is strictly less than that of M. Finally swap digits if valid.

``````class Solution {
func maximumSwap(_ num: Int) -> Int {
var M = -1
let s = String(num)
var L = s.characters.count
var R = -1
var curM = -1
var nums = s.characters.map {
Int(String(\$0))!
}
for (i, d) in nums.enumerated().reversed() {
if d < M {
L = i
R = curM
}
else if d > M {
M = d
curM = i
}
}
guard L < nums.count && R > -1 else {
return num
}
let t = nums[L]
nums[L] = nums[R]
nums[R] = t
return Int(nums.map{String(\$0)}.joined(separator: ""))!
}
}
``````

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