diff --git a/complexity Analysis/Dante's solutions/largest_secondsmallest/largest.js b/complexity Analysis/Dante's solutions/largest_secondsmallest/largest.js
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+++ b/complexity Analysis/Dante's solutions/largest_secondsmallest/largest.js	
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+/**
+ * First Problem Statement: Given an array, we have to find the largest element in the array.
+
+ 
+
+Examples
+
+Plain Text
+Example 1:Input: arr[] = {2,5,1,3,0};Output: 5Explanation: 5 is the largest element in the array. 
+Example2:Input: arr[] = {8,10,5,7,9};Output: 10Explanation: 10 is the largest element in the array.
+ */
+
+
+function findLargest(arr){
+
+    if(arr.length===0){
+        return -1
+    }else{
+ return Math.max(...arr);
+    }
+}
+
+//test
+let array=[2,5,1,3,0]
+let num=findLargest(array)
+console.log(num)
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diff --git a/complexity Analysis/Dante's solutions/largest_secondsmallest/secondsmallest.js b/complexity Analysis/Dante's solutions/largest_secondsmallest/secondsmallest.js
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+++ b/complexity Analysis/Dante's solutions/largest_secondsmallest/secondsmallest.js	
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+
+/**
+ * 
+ *  Statement: Given an array, find the second smallest in the array. Print ‘-1’ in the event that it doesn’t exist.
+
+ 
+
+Examples
+
+ 
+
+Plain Text
+Example 1:Input: [1,2,4,7,7,5]Output: Second Smallest : 2
+
+  Second smallest : 2Explanation: The elements are as follows 1,2,3,5,7,7 and hence  second smallest is 2
+
+Example 2:Input: [1]Output: Second Smallest : -1
+  Second Largest : -1Explanation: Since there is only one element in the array, it is the smallest element present in the array. There is no second smallest element present.in the arra
+ */
+
+
+
+function secondSmallest(arr){
+    let minimum=Math.min(...arr)
+    let minIndex=arr.indexOf(minimum)
+    if(arr.length==0||arr.length==1){
+        return -1
+    }
+    else{
+       arr.splice(minIndex,1) 
+       return Math.min(...arr)
+        
+
+
+    }
+}
+
+let array=[1,2,4,7,7,5]
+let newArray=secondSmallest(array)
+console.log(newArray);
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