feat: add Find the Smallest Divisor Given a Threshold (#1175)

* add leetcode Find the Smallest Divisor Given a Threshold

* Update leetcode/DIRECTORY.md

Co-authored-by: David Leal <halfpacho@gmail.com>

leetcode/DIRECTORY.md

@@ -113,6 +113,7 @@
 | 1184 | [Distance Between Bus Stops](https://leetcode.com/problems/distance-between-bus-stops/)                                         | [C](./src/1184.c) | Easy       |
 | 1189 | [Maximum Number of Balloons](https://leetcode.com/problems/maximum-number-of-balloons/)                                         | [C](./src/1189.c) | Easy       |
 | 1207 | [Unique Number of Occurrences](https://leetcode.com/problems/unique-number-of-occurrences/)                                     | [C](./src/1207.c) | Easy       |
+| 1283 | [Find the Smallest Divisor Given a Threshold]https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/description/)  | [C](./src/1283.c) | Medium       |
 | 1695 | [Maximum Erasure Value](https://leetcode.com/problems/maximum-erasure-value/)                                                   | [C](./src/1695.c) | Medium     |
 | 1769 | [Minimum Number of Operations to Move All Balls to Each Box](https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/)          | [C](./src/1769.c) | Medium     |
 | 1524 | [Number of Sub-arrays With Odd Sum](https://leetcode.com/problems/number-of-sub-arrays-with-odd-sum/)                           | [C](./src/1524.c) | Medium     |

leetcode/src/1283.c

@@ -0,0 +1,44 @@
+#define max(a,b) (((a)>(b))?(a):(b))
+
+long getSum(int* nums, int numsSize, int divizor){
+    long result = 0;
+    for (int i = 0; i < numsSize; i++){
+        int value = nums[i] / divizor;
+        if (value * divizor != nums[i]){
+            value++;
+        }
+
+        result += value;
+    }
+
+    return result;
+}
+
+// Divide and conquer
+// Runtime: O(n*log(n))
+// Space: O(1)
+int smallestDivisor(int* nums, int numsSize, int threshold){
+    int maxNum = 0;
+    for (int i = 0; i < numsSize; i++){
+        maxNum = max(maxNum, nums[i]);
+    }
+
+    int left = 1;
+    int right = maxNum;
+    while (left <= right){
+        int middle = (left + right) / 2;
+        long middleSum = getSum(nums, numsSize, middle);
+        if (middleSum <= threshold && (middle == 1 || getSum(nums, numsSize, middle - 1) > threshold)){
+            return middle;
+        }
+
+        if (middleSum > threshold){
+            left = middle + 1;
+        }
+        else{
+            right = middle - 1;
+        }
+    }
+
+    return -1;
+}