Merge pull request #297 from nik132-eng/patch-7

Create Max_Chunks_To_Make_Sorted_II.java

Author: SarthakKeshari

LeetCode/Max_Chunks_To_Make_Sorted_II.java

@@ -0,0 +1,74 @@
+
+//768. Max Chunks To Make Sorted II
+
+//You are given an integer array arr.
+
+//We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.
+
+//Return the largest number of chunks we can make to sort the array.
+ 
+
+//Example 1:
+
+//Input: arr = [5,4,3,2,1]
+//Output: 1
+//Explanation:
+//Splitting into two or more chunks will not return the required result.
+//For example, splitting into [5, 4], [3, 2, 1] will result in [4, 5, 1, 2, 3], which isn't sorted.
+
+//Example 2:
+
+//Input: arr = [2,1,3,4,4]
+//Output: 4
+//Explanation:
+//We can split into two chunks, such as [2, 1], [3, 4, 4].
+//However, splitting into [2, 1], [3], [4], [4] is the highest number of chunks possible.
+ 
+
+//Constraints:
+
+//1 <= arr.length <= 2000
+//0 <= arr[i] <= 108
+
+import java.util.Scanner;
+
+public class Max_Chunks_To_Make_Sorted_II {
+
+	public static void main(String[] args) {
+		Scanner sc = new Scanner(System.in);
+
+		System.out.println("Enter Number of element : ");
+		int n = sc.nextInt();
+
+		int arr[] = new int[n];
+		System.out.println("Enter " + n + " values :");
+		for (int i = 0; i < n; i++) {
+			arr[i] = sc.nextInt();
+		}
+		int ans = maxChunksToSorted2(arr);
+
+		System.out.println("Maximun chunk to sorts are : " + ans);
+	}
+
+	private static int maxChunksToSorted2(int[] arr) {
+//		generate right minimum 
+		int[] rmin = new int[arr.length + 1];
+
+		rmin[arr.length] = Integer.MAX_VALUE;
+		for (int i = arr.length - 1; i >= 0; i--) {
+			rmin[i] = Math.min(rmin[i + 1], arr[i]);
+		}
+
+//		iterate on array and manage left maximum as well as count chunk
+		int lmax = Integer.MIN_VALUE;
+		int count = 0;
+		for (int i = 0; i < arr.length; i++) {
+			lmax = Math.max(lmax, arr[i]);
+//			if the left max element is less than right minimum element than left partis one of the chunk so increment count
+			if (lmax <= rmin[i + 1])
+				count++;
+		}
+		return count;
+	}
+
+}