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Implemented 'Next Greater Element III' challenge
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| Original file line number | Diff line number | Diff line change |
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| // | ||
| // NextGreaterElementIII.swift | ||
| // LeetCode | ||
| // | ||
| // Created by William Boles on 05/09/2024. | ||
| // | ||
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| import Foundation | ||
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| //https://leetcode.com/problems/next-greater-element-iii/ | ||
| struct NextGreaterElementIII { | ||
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| //Time: O(n) where n is the number of elements in `nums` | ||
| //Space: O(n) | ||
| //array | ||
| //two pointers | ||
| //inline | ||
| //permutation | ||
| //pivot & swap | ||
| // | ||
| //Solution Description: | ||
| //The next greater element is the next permutation of `n` that is the smallest version that is greater than `n`. To find | ||
| //this "smallest greater than" permutation we need to find an element that when swapped with another minimally increases | ||
| //`n` in value. The element that minimally increases `n` in value is the right-most element that when reading from right | ||
| //to left is smaller than it's right neighbor i.e. the elements to it's right are in ascending order and it is the break | ||
| //element that goes in descending order e.g. | ||
| // | ||
| //In `834762` the `4` is that value | ||
| // | ||
| //We call this breaking pattern element the `pivot`. Once we have the `pivot` we need to find the smallest value that is | ||
| //greater than it to swap over - this is to minimise the increase in `n`. You might be thinking if everything to the | ||
| //right of `pivot` is ascending won't that smallest value be the last digit in `n` - be careful here, just because the | ||
| //`pivot` is smaller than it's neighor doesn't mean that it is smaller than all elements to the right and swapping the | ||
| //`pivot` with an element smaller than itself would not result in the smallest permutation of `n` greater than `n` but | ||
| //rather a smaller permutation of `n` e.g. | ||
| // | ||
| //In `834762` we don't want to swap the `4` with the `2` as `832764` < `834762` instead we want to swap it with the `6` | ||
| // | ||
| //So having swapped the `pivot` with the smallest right neighbor that is still greater than the `pivot` e.g. | ||
| // | ||
| //`836742` | ||
| // | ||
| //we can still make this number smaller by reversing those ascending numbers to the right of the old `pivot` index to | ||
| //descending order e.g. | ||
| // | ||
| //`836247` | ||
| // | ||
| //With our smallest permutation we just need to check that it isn't larger than Int32.max and return either it or -1. | ||
| // | ||
| //Similar to https://leetcode.com/problems/next-permutation/ | ||
| func nextGreaterElement(_ n: Int) -> Int { | ||
| guard n > 9 else { | ||
| return -1 | ||
| } | ||
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| var nums = Array(String(n)) | ||
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| //find the least significant value that is smaller than it's right neighbor | ||
| var pivot = nums.count - 2 | ||
| while pivot >= 0, nums[(pivot + 1)] <= nums[pivot] { | ||
| pivot -= 1 | ||
| } | ||
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| guard pivot >= 0 else { | ||
| //`n` is already as large as it can be | ||
| return -1 | ||
| } | ||
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| //find the least-significant-value to the right of the pivot that is larger than the pivot as we want | ||
| //`nums` to be > `n` | ||
| var lsv = nums.count - 1 | ||
| while lsv > 0, nums[lsv] <= nums[pivot] { | ||
| lsv -= 1 | ||
| } | ||
| nums.swapAt(pivot, lsv) | ||
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| //now the pivot is moved into place making `nums` > `n` we need to minimise `nums` so it's as small | ||
| //as can be while still > `n` | ||
| nums[(pivot + 1)...].reverse() | ||
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| guard let num32 = Int32(String(nums)) else { //casting to Int32 as problem expects Int32 | ||
| return -1 | ||
| } | ||
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| let num = Int(num32) //casting to Int to make comparison easier with `n` | ||
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| return num > n ? num : -1 | ||
| } | ||
| } |
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47 changes: 47 additions & 0 deletions
47
LeetCode/LeetCodeTests/Tests/NextGreaterElementIIITests.swift
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,47 @@ | ||
| // | ||
| // NextGreaterElementIIITests.swift | ||
| // LeetCodeTests | ||
| // | ||
| // Created by William Boles on 05/09/2024. | ||
| // | ||
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| import XCTest | ||
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| @testable import LeetCode | ||
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| final class NextGreaterElementIIITests: XCTestCase { | ||
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| // MARK: - Tests | ||
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| func test_A() { | ||
| let n = 12 | ||
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| let result = NextGreaterElementIII().nextGreaterElement(n) | ||
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| XCTAssertEqual(result, 21) | ||
| } | ||
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| func test_B() { | ||
| let n = 21 | ||
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| let result = NextGreaterElementIII().nextGreaterElement(n) | ||
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| XCTAssertEqual(result, -1) | ||
| } | ||
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| func test_C() { | ||
| let n = 2147483486 | ||
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| let result = NextGreaterElementIII().nextGreaterElement(n) | ||
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| XCTAssertEqual(result, -1) | ||
| } | ||
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| func test_D() { | ||
| let n = 2147483476 | ||
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| let result = NextGreaterElementIII().nextGreaterElement(n) | ||
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| XCTAssertEqual(result, 2147483647) | ||
| } | ||
| } |
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