chore(Java): add next greater element (#833)

* chore(Java): added Next Greater Element

* Update Next_Greater_Element.java file

algorithms/Java/README.md

@@ -75,6 +75,7 @@
 - [Celebrity Problem](stacks/celebrity-problem.java)
 - [Sliding Window Maximum](stacks/sliding-window-maximum.java)
 - [Min Stack](stacks/Min-Stack.java)
+- [Next Greater Element](stacks/Next_Greater_Element.java)
 
 ## Strings
 

algorithms/Java/stacks/Next_Greater_Element.java

@@ -0,0 +1,79 @@
+/***
+ * Problem Statement:
+ * Given an array arr[] of size N having distinct elements, the task is to find the next greater element for each element of the array in order of their appearance in the array.
+ * Next greater element means nearest element on right side which is greater than current element, 
+ * if there is no suxh element, then put -1 for that.
+ *
+ */
+
+
+ /***
+  * Example 1: Input: arr=[1,3,2,4] n=4
+               Output: 3 4 4 -1
+    
+    Example 2: Input: arr=[6 8 0 1 3] n=5
+               Output: 8 -1 1 3 -1
+  */
+
+  /***
+   * Time Complexity: O(N)
+   * Space Complexity: O(N)
+   */
+  import java.util.Scanner;
+import java.util.Stack;
+  
+  public class Next_Greater_Element {
+      public static void main(String[] args){
+          Scanner input=new Scanner(System.in);
+          int n=input.nextInt();
+          long arr[]=new long[n];
+          for (int i=0;i<n;i++){
+              arr[i]=input.nextLong();
+          }
+          long ans[]=nextLargerElement(arr,n);  // functions that print array for calculating next greater element
+          for (int i=0;i<ans.length;i++){
+            System.out.print(ans[i]+" ")
+          }
+    }
+      public static long[] nextLargerElement(long[] arr, int n)
+      { 
+          // Your code here
+
+        /****
+         * Approach: for element at index i,
+         * there were two posiibilities that there is greater element than this
+         * at right side of array or not.
+         * if there is no greater element in right side is indicated by Empty stack
+         * if there is greater ekement in right side is indicated by stack peek element.
+         */
+
+
+          long ans[]=new long[n];
+          ans[n-1]=-1;
+          Stack<Long> st=new Stack<>();
+          st.push(arr[n-1]);
+          for (int i=n-2;i>=0;i--)
+          {
+              if(st.isEmpty())
+              {
+                  st.push(arr[i]);
+                  ans[i]=-1;
+              }
+              else
+              {
+                  while(st.isEmpty()==false && arr[i]>=st.peek())
+                  {
+                      st.pop();
+                  }
+                  if(st.size()==0){
+                      ans[i]=-1;
+                  }else{
+                      ans[i]=st.peek();
+                  }
+                  st.push(arr[i]);
+              }
+          }
+          return ans;
+      } 
+  }
+