Adds 1283. Find the Smallest Divisor Given a Threshold

Author: ivanlesko

1283. Find the Smallest Divisor Given a Threshold .playground/Contents.swift

@@ -0,0 +1,63 @@
+import UIKit
+
+/*
+ # Intuition
+ Define a solution space to perform a binary search to find the min # of banans.
+ 
+ # Approach
+ Define a range of possible values.  We know that the mininum possible value
+ is 1 since dividing a number by 0 won't accumulate the sum.  The maximum possible
+ value is `max(nums)` since any greater than the max will always result in 1.
+ Use binary search to half the possible range of answer on each iteration.
+ The sum of values increases as the divisor decreases and
+ the sum of values decreases as the divisor increases.
+ 
+ Continue reducing the binary search space until it cannot be reduced further.
+ 
+ # Complexity
+ - Time complexity:
+ O(n log k)
+ O(n)(check) * O(log k)(binary search where `k` = `nums.max`)
+ 
+ - Space complexity:
+ O(1) only a few integers are stored.
+ 
+ This problem is very similar to Koko eating bananas: https://leetcode.com/problems/koko-eating-bananas/
+ */
+
+class Solution {
+    func check(_ nums: [Int], _ threshold: Int, _ divisor: Int) -> Bool {
+        var sum = 0
+        for num in nums {
+            sum += Int(ceil(Double(num) / Double(divisor)))
+            if sum > threshold {
+                return false
+            }
+        }
+
+        return sum <= threshold
+    }
+
+    func smallestDivisor(_ nums: [Int], _ threshold: Int) -> Int {
+        var l = 1, r = Int.min
+        for num in nums {
+            r = max(num, r)
+        }
+
+        while l <= r {
+            let mid = (l + r) / 2
+            if check(nums, threshold, mid) {
+                // sum <= threshold. Decrease divisor to increase sum.
+                r = mid - 1
+            } else {
+                // sum > threshold.  Increase divisor to decrease sum.
+                l = mid + 1
+            }
+        }
+
+        return l
+    }
+}
+
+let res1 = Solution().smallestDivisor([1,2,5,9], 6) // 5
+let res2 = Solution().smallestDivisor([44,22,33,11,1], 5) // 44

1283. Find the Smallest Divisor Given a Threshold .playground/contents.xcplayground

@@ -0,0 +1,4 @@
+<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
+<playground version='5.0' target-platform='ios' buildActiveScheme='true' importAppTypes='true'>
+    <timeline fileName='timeline.xctimeline'/>
+</playground>