update 852 java

Author: yennj12

data/progress.md

@@ -2,6 +2,7 @@
 
 # 2025-01-08
 - https://github.com/yennanliu/CS_basics/blob/master/doc/Leetcode_company_frequency-master/Google%206months-%20LeetCode.pdf
+- TODO: go through `neetcode`, note/summary knowledges (by topic) at cheatsheet
 
 # 2025-01-07
 - https://github.com/yennanliu/CS_basics/blob/master/doc/Leetcode_company_frequency-master/Google%206months-%20LeetCode.pdf

leetcode_java/src/main/java/LeetCodeJava/BinarySearch/PeakIndexInAMountainArray.java

@@ -61,9 +61,82 @@
 public class PeakIndexInAMountainArray {
 
     // V0
+    // TODO: fix below:
 //    public int peakIndexInMountainArray(int[] arr) {
 //
+//        // edge
+//        int maxIdx = -1;
+//        int maxVal = -1;
+//        if (arr.length == 3){
+//            for(int i = 0; i < arr.length; i++){
+//                if(arr[i] > maxVal){
+//                    maxVal = arr[i];
+//                    maxIdx = i;
+//                }
+//            }
+//            return maxIdx;
+//        }
+//
+//        // binary search
+//        int l = 0;
+//        int r = arr.length - 1;
+//        int mid = (l + r) / 2;
+//        while (r >= l && r >= 0){
+//
+//            mid = (l + r) / 2;
+//
+//            // case 1)  cur > left and cur > right (find peak)
+//            if (arr[mid] > arr[mid-1] && arr[mid] > arr[mid+1]){
+//                return mid;
+//            }
+//            // Exp 1 : [0,0,0, 3,2,1,0] -> 1
+//            // case 2) cur < left && cur < left most (left is increasing order)
+//            else if (arr[mid] >= arr[mid-1] && arr[mid] >= arr[l]){
+//                l = mid + 1;
+//            }
+//            // case 3) cur < right and cur > right most (right is decreasing order ?)
+//            else if  (arr[mid] >= arr[mid+1] && arr[mid] >= arr[r]){
+//                r = mid - 1;
+//            }
+//        }
+//
+//        return mid;
 //    }
+
+    // V0-1
+    // IDEA: BINARY SEARCH
+    // TODO: validate & fix
+    // https://github.com/yennanliu/CS_basics/blob/master/leetcode_python/Binary_Search/peak-index-in-a-mountain-array.py#L55
+    public int peakIndexInMountainArray_0_1(int[] arr) {
+        if (arr.length < 3) {
+            return -1; // Return -1 if the array length is less than 3
+        }
+
+        // Binary search
+        int l = 0;
+        int r = arr.length - 1;
+
+        while (r >= l) {
+            int mid = l + (r - l) / 2;
+
+            // Check if mid is the peak
+            if (arr[mid] > arr[mid - 1] && arr[mid] > arr[mid + 1]) {
+                return mid;
+            }
+            // If the element at mid is smaller than the next element, peak is on the right
+            else if (arr[mid] < arr[mid + 1]) {
+                l = mid + 1;
+            }
+            // Otherwise, peak is on the left
+            else {
+                r = mid - 1;
+            }
+        }
+        return -1; // Return -1 if no peak is found (though this case shouldn't happen with a valid
+        // mountain array)
+    }
+
     // V1
+
     // V2
 }

leetcode_java/src/main/java/dev/workspace6.java

@@ -2246,6 +2246,16 @@ private boolean isStrobogrammatic(String x){
 
     // LC 852
     // https://leetcode.com/problems/peak-index-in-a-mountain-array/
+    /**
+     * You are given an integer mountain array arr
+     * of length n where the values increase to a peak element and then decrease.
+     *
+     * Return the index of the peak element.
+     *
+     * Your task is to solve it in O(log(n)) time complexity.
+     *
+     *
+     */
     /**
      *
      *
@@ -2293,27 +2303,27 @@ public int peakIndexInMountainArray(int[] arr) {
         // binary search
         int l = 0;
         int r = arr.length - 1;
-        while (r >= l){
-            int mid = (l + r) / 2;
+        int mid = (l + r) / 2;
+        while (r >= l && r >= 0){
+
+            mid = (l + r) / 2;
 
             // case 1)  cur > left and cur > right (find peak)
             if (arr[mid] > arr[mid-1] && arr[mid] > arr[mid+1]){
                 return mid;
             }
             // Exp 1 : [0,0,0, 3,2,1,0] -> 1
-            // case 2) cur < left && cur > left most
-            else if (arr[mid] < arr[mid-1] && arr[mid] >= arr[l]){
+            // case 2) cur < left && cur < left most (left is increasing order)
+            else if (arr[mid] >= arr[mid-1] && arr[mid] >= arr[l]){
                 l = mid + 1;
             }
-            // case 3) cur < right and cur > right most
-            else if  (arr[mid] < arr[mid-1] && arr[mid] >= arr[r]){
+            // case 3) cur < right and cur > right most (right is decreasing order ?)
+            else if  (arr[mid] >= arr[mid+1] && arr[mid] >= arr[r]){
                 r = mid - 1;
         }
-
-        return 0;
     }
 
-        return -1;
+        return mid;
     }
 
 }