✨ [153] Watched solution

Author: enkr1

153/my_solution.js

@@ -2,40 +2,51 @@
  * @param {number[]} nums
  * @return {number}
  */
-const findMin = (nums) => {
+const findMin = (nums) => { // O(log n)
   // if (nums[0] < nums[nums.length - 1]) {
   //   console.log("nums.length - 1")
   //   console.log(nums.length - 1)
   //   return nums.length - 1;
   // }
-
-  let l = 0, r = nums.length - 1, count = 0, min = 0;
+  // let p = Math.ceil(nums.length / 2) - 1, l = 0, r = nums.length - 1, min = nums[p];
+  let l = 0, r = nums.length - 1, min = nums[0];
   // console.log(`l:${l}, r:${r}`)
 
-  while (l < r) {
-    console.log(`l:${l}, r:${r}`)
+  while (l <= r) {
     if (nums[l] < nums[r]) {
+      min = Math.min(min, nums[l])
       break;
     }
 
-    count++;
+    let p = Math.ceil((r + l) / 2);
+
+    console.log(`l:${l}, r:${r}, p:${p}, nums[p]:${nums[p]}`)
+    // L <= mid = they are the same chunk
+    min = Math.min(min, nums[p]);
 
-    if (nums[l] > nums[r]) {
-      l++;
-      r--;
+    if (nums[l] <= nums[p]) {
+      // Set L to the most right.
+      l = p + 1;
+    } else {
+      // L > Mid, mid is in the R chunk
+      r = p - 1;
     }
+    // // Find the mid of the new chunk
+    // p = Math.ceil(((r - l) / 2) + l) - 1;
   }
 
-  count++;
-
-  console.log("count")
-  console.log(count)
+  console.log("min")
+  console.log(min)
 
   // if l==0 && l < r , count = nums.length
-  return count;
+  return min;
 };
 
 // findMin([3, 4, 5, 1, 2]) // 3
 // findMin([4, 5, 6, 7, 0, 1, 2]) // 4
-findMin([11, 13, 15, 17]) // 4
+// findMin([11, 13, 15, 17]) // 4
 // findMin([3, 4, 5, 6, 7, 0, 1, 2]) // 1
+// findMin([1]) // 1
+// findMin([3, 1, 2]) // 1
+
+findMin([2, 1]) // 1

153/solution.js

@@ -0,0 +1,26 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const findMin = (nums) => {
+  let l = 0, r = nums.length - 1, min = nums[0];
+
+  while (l <= r) {
+    if (nums[l] < nums[r]) {
+      min = Math.min(min, nums[l])
+      break;
+    }
+
+    let p = Math.ceil((r + l) / 2);
+
+    min = Math.min(min, nums[p]);
+
+    if (nums[l] <= nums[p]) {
+      l = p + 1;
+    } else {
+      r = p - 1;
+    }
+  }
+
+  return min;
+};