Merge pull request #33 from abhishektripathi66/abhishektripathi66-pat…

…ch-1

Create MinimumNumberofRemovalstoMakeMountainArray.java

Leetcode/MinimumNumberofRemovalstoMakeMountainArray.java

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+/**
+1671. Minimum Number of Removals to Make Mountain Array
+Solved
+Hard
+Topics
+Companies
+Hint
+You may recall that an array arr is a mountain array if and only if:
+
+arr.length >= 3
+There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that:
+arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
+arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
+Given an integer array nums​​​, return the minimum number of elements to remove to make nums​​​ a mountain array.
+  **/
+class Solution {
+    public int minimumMountainRemovals(int[] nums) {
+        int n = nums.length;
+        int[] LIS = new int[n], LDS = new int[n];
+        Arrays.fill(LIS, 1);
+        Arrays.fill(LDS, 1);
+
+        // Compute LIS for each index
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < i; ++j) {
+                if (nums[i] > nums[j]) {
+                    LIS[i] = Math.max(LIS[i], LIS[j] + 1);
+                }
+            }
+        }
+
+        // Compute LDS from each index
+        for (int i = n - 1; i >= 0; --i) {
+            for (int j = n - 1; j > i; --j) {
+                if (nums[i] > nums[j]) {
+                    LDS[i] = Math.max(LDS[i], LDS[j] + 1);
+                }
+            }
+        }
+
+        int maxMountainLength = 0;
+
+        // Find the maximum mountain length
+        for (int i = 1; i < n - 1; ++i) {
+            if (LIS[i] > 1 && LDS[i] > 1) {  // Valid peak
+                maxMountainLength = Math.max(maxMountainLength, LIS[i] + LDS[i] - 1);
+            }
+        }
+
+        return n - maxMountainLength;
+    }
+}