From dbfc2a6a551d685bd38a2ed1f00fb100acacdf8d Mon Sep 17 00:00:00 2001
From: Aneesh Gupta <77202061+aneeshgupta25@users.noreply.github.com>
Date: Mon, 3 Oct 2022 02:28:30 +0530
Subject: [PATCH] Adding solution for Merge Intervals (Leetcode 56)

---
 Java/Stack/Merge_Intervals.java | 38 +++++++++++++++++++++++++++++++++
 1 file changed, 38 insertions(+)
 create mode 100644 Java/Stack/Merge_Intervals.java

diff --git a/Java/Stack/Merge_Intervals.java b/Java/Stack/Merge_Intervals.java
new file mode 100644
index 00000000..99d4d6f1
--- /dev/null
+++ b/Java/Stack/Merge_Intervals.java
@@ -0,0 +1,38 @@
+/*Problem
+56. Merge Intervals (LEETCODE)
+Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping 
+intervals that cover all the intervals in the input.
+
+Meaning -> For example, if we have 2 meetings, one is scheduled from 7am-8am and the other one is from 7:30am-9am. Then we can merge these
+2 meetings 2 get 1 -> (7am-9am).
+
+How to implement -> We'll be using Stack data structure to solve this problem.
+
+IDEOLOGY -> Sort the given elements of the array. Then merge 2 meetings if start time of the second meeting is encountered before the end time of the first meeting. */
+
+
+class Solution {
+    public int[][] merge(int[][] intervals) {
+        Arrays.sort(intervals, (a,b)-> Integer.compare(a[0], b[0]));    //This sorting will sort the meetings according to their start time.
+
+        Stack<int[]> st = new Stack<>();
+        st.push(intervals[0]);    //Initially, we'll push the first meeting element into the stack.
+
+        for(int i = 1; i < intervals.length; i++){
+            if(st.peek()[1] >= intervals[i][0]){
+                int[] arr = st.pop();
+                arr[1] = Math.max(arr[1], intervals[i][1]);
+                st.push(arr);
+            } else {
+                st.push(intervals[i]);
+            }
+        }
+        int[][] ans = new int[st.size()][2];
+        int index = ans.length-1;
+        while(index >= 0){
+            ans[index] = st.pop();
+            index--;
+        }
+        return ans;
+    }
+}