chore(Java): add celebrity problem and sliding window maximum (#281)

algorithms/Java/README.md

@@ -53,6 +53,8 @@
 1. [Balanced Parenthesis](stacks/balanced-paranthesis.java)
 2. [Stack](stacks/stack.java)
 3. [The Stock Span Problem](stacks/the-stock-span-problem.java)
+4. [Celebrity Problem](stacks/celebrity-problem.java)
+5. [Sliding Window Maximum](stacks/sliding-window-maximum.java)
 
 ## Strings
 

algorithms/Java/stacks/celebrity-problem.java

@@ -0,0 +1,78 @@
+//Problem Statement
+// 1. You are given a number n, representing the number of people in a party.
+// 2. You are given n strings of n length containing 0's and 1's
+// 3. If there is a '1' in ith row, jth spot, then person i knows about person j.
+// 4. A celebrity is defined as somebody who knows no other person than himself but everybody else knows him.
+// 5. If there is a celebrity print it's index otherwise print "none".
+
+//solution
+
+import java.io.*;
+import java.util.*;
+
+public class Main {
+
+    public static void main(String[] args) throws Exception {
+        // write your code here
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        int n = Integer.parseInt(br.readLine());
+        int[][] arr = new int[n][n];
+
+        for (int j = 0; j < n; j++) {
+            String line = br.readLine();
+            for (int k = 0; k < n; k++) {
+                arr[j][k] = line.charAt(k) - '0';
+            }
+        }
+
+        findCelebrity(arr);
+    }
+
+    public static void findCelebrity(int[][] arr) {
+        // if a celebrity is there print it's index (not position), if there is not then print "none"
+        Stack < Integer > st = new Stack < > ();
+        for (int i = 0; i < arr.length; i++) {
+            st.push(i);
+        }
+
+        while (st.size() > 1) {
+            int i = st.pop();
+            int j = st.pop();
+
+            if (arr[i][j] == 1) {
+                st.push(j);
+            } else {
+                st.push(i);
+            }
+        }
+
+        int pot = st.pop();
+        boolean flag = true;
+        for (int i = 0; i < arr.length; i++) {
+            if (i != pot) {
+                if (arr[i][pot] == 0 || arr[pot][i] == 1) {
+                    flag = false;
+                    break;
+                }
+            }
+        }
+
+        if (flag) {
+            System.out.println(pot);
+        } else {
+            System.out.println("none");
+        }
+    }
+
+}
+
+//Test Case:-
+//Input
+// 4
+// 0000
+// 1011
+// 1101
+// 1110
+
+//Output
+//0

algorithms/Java/stacks/sliding-window-maximum.java

@@ -0,0 +1,65 @@
+//Statement
+// 1. You are given a number n, representing the size of array a.
+// 2. You are given n numbers, representing the elements of array a.
+// 3. You are given a number k, representing the size of window.
+// 4. You are required to find and print the maximum element in every window of size k.
+
+// e.g.
+// for the array [2 9 3 8 1 7 12 6 14 4 32 0 7 19 8 12 6] and k = 4, the answer is [9 9 8 12 12 14 14 32 32 32 32 19 19 19]
+
+//solution
+import java.io.*;
+import java.util.*;
+
+public class Main{
+  
+
+public static void main(String[] args) throws Exception {
+    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+
+    int n = Integer.parseInt(br.readLine());
+    int[] arr = new int[n];
+    for(int i = 0; i < n; i++){
+       arr[i] = Integer.parseInt(br.readLine());
+    }
+    int k = Integer.parseInt(br.readLine());
+
+    // code
+	//nge is next Greater Element
+    // nge begin 
+    int[] nge = new int[arr.length];
+
+    Stack<Integer> st = new Stack<>();
+    st.push(arr.length - 1);
+    nge[arr.length - 1] = arr.length;
+
+    for(int i = arr.length - 2; i >= 0; i--){
+      while(st.size() > 0 && arr[i] >= arr[st.peek()]){
+        st.pop();
+      }
+
+      if(st.size() == 0){
+        nge[i] = arr.length;
+      } else {
+        nge[i] = st.peek();
+      }
+
+      st.push(i);
+    }
+
+    // nge end
+
+    int j = 0;
+    for(int i = 0; i <= arr.length - k; i++){
+      if(j < i){
+        j = i;
+      }
+
+      while(nge[j] < i + k){
+        j = nge[j];
+      }
+      System.out.println(arr[j]);
+    }
+ }
+}
+