Stack1 completed. Please review

DailyTemperatures.java

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+package Stack1;
+
+public class DailyTemperatures {
+	//TC : o(2n) - o(n) to read each temperature and push to stack and o(n) to resolve each from stack. So o(2n)
+	//SC : o(n) no. of elements in array
+	//Approach : Brute force way is to nested iterations to check next largest element which o(n^2) time
+	// Optimized : push each index to stack and check if arr[i] is greater than top of stack. If it is then calculate value at that index(i-index) 
+	class Solution {
+	    public int[] dailyTemperatures(int[] temperatures) {
+	        //null case
+	        int[] result = new int[temperatures.length];
+	        if(temperatures == null || temperatures.length == 0) return result;
+
+	        Stack<Integer> st = new Stack<>();
+
+	        for(int i=0; i<temperatures.length; i++){
+	            while(!st.isEmpty() && temperatures[i] > temperatures[st.peek()]){
+	                int index = st.pop();
+	                result[index] = i - index;
+	            }
+	                st.push(i);
+	        }
+	        return result;
+	    }
+	}
+}

NextGreaterElement2.java

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+package Stack1;
+
+public class NextGreaterElement2 {
+	//TC : o(4n)
+	//SC : o(n)
+	//Approach : fill result array with -1's. push each index into stack and check nums[i] > stack.peek() then add value of that index to res array. array can be iterated twice as it's circular
+	class Solution {
+	    public int[] nextGreaterElements(int[] nums) {
+	        int[] res = new int[nums.length];
+	        if(nums == null || nums.length == 0) return res;
+	        Stack<Integer> st = new Stack<>();
+	        Arrays.fill(res,-1);
+	        int n = nums.length;
+	        for(int i=0; i<2*n; i++){
+	            while(!st.isEmpty() && nums[i%n] > nums[st.peek()]){
+	                int index = st.pop();
+	                res[index] = nums[i%n];
+	            }
+	            if(i<n) st.push(i);
+	        }
+	        return res;
+	    }
+	}
+}