diff --git a/C++/Kadane's_Algorithm.cpp b/C++/Kadane's_Algorithm.cpp
new file mode 100644
index 0000000..1c84dbc
--- /dev/null
+++ b/C++/Kadane's_Algorithm.cpp
@@ -0,0 +1,31 @@
+// Problem Statement 
+/*
+Write an efficient program to find the sum of contiguous subarray within a one-dimensional array 
+of numbers which has the largest sum.
+*/
+
+#include<iostream> 
+using namespace std; 
+
+int maxSubArraySum(int a[], int size) 
+{ 
+int max_so_far = a[0]; 
+int curr_max = a[0]; 
+
+for (int i = 1; i < size; i++) 
+{ 
+		curr_max = max(a[i], curr_max+a[i]); 
+		max_so_far = max(max_so_far, curr_max); 
+} 
+return max_so_far; 
+} 
+
+/* Driver program to test maxSubArraySum */
+int main() 
+{ 
+int a[] = {-2, -3, 4, -1, -2, 1, 5, -3}; 
+int n = sizeof(a)/sizeof(a[0]); 
+int max_sum = maxSubArraySum(a, n); 
+cout << "Maximum contiguous sum is " << max_sum; 
+return 0; 
+} 
diff --git a/C++/Knapsack_0-1.cpp b/C++/Knapsack_0-1.cpp
new file mode 100644
index 0000000..42374a5
--- /dev/null
+++ b/C++/Knapsack_0-1.cpp
@@ -0,0 +1,52 @@
+// Problem Statement 
+/*
+Given weights and values of n items, put these items in a knapsack of capacity W to
+get the maximum total value in the knapsack. In other words, given two integer arrays
+val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items
+respectively. Also given an integer W which represents knapsack capacity, 
+find out the maximum value subset of val[] such that sum of the weights 
+of this subset is smaller than or equal to W. You cannot break an item, 
+either pick the complete item or don’t pick it (0-1 property).
+
+*/
+
+// Algorithm :
+
+// A Dynamic Programming based 
+// solution for 0-1 Knapsack problem 
+#include<bits/stdc++.h>
+using namespace std ;
+
+// Returns the maximum value that 
+// can be put in a knapsack of capacity W 
+int knapSack(int W, int wt[], int val[], int n) 
+{ 
+	int i, w; 
+	int K[n + 1][W + 1]; 
+
+	// Build table K[][] in bottom up manner 
+	for (i = 0; i <= n; i++) { 
+		for (w = 0; w <= W; w++) { 
+			if (i == 0 || w == 0) 
+				K[i][w] = 0; 
+			else if (wt[i - 1] <= w) 
+				K[i][w] = max( 
+					val[i - 1] + K[i - 1][w - wt[i - 1]], 
+					K[i - 1][w]); 
+			else
+				K[i][w] = K[i - 1][w]; 
+		} 
+	} 
+
+	return K[n][W]; 
+} 
+
+int main() 
+{ 
+	int val[] = { 60, 100, 120 }; 
+	int wt[] = { 10, 20, 30 }; 
+	int W = 50; 
+	int n = sizeof(val) / sizeof(val[0]); 
+	out<< knapSack(W, wt, val, n)) <<'\n'; 
+	return 0; 
+} 
diff --git a/C++/Longest_Common_Subsequence.cpp b/C++/Longest_Common_Subsequence.cpp
new file mode 100644
index 0000000..5f170e8
--- /dev/null
+++ b/C++/Longest_Common_Subsequence.cpp
@@ -0,0 +1,38 @@
+#include<bits/stdc++.h>
+using namespace std ;
+
+int LCS(string a,string b){
+    int m = a.length() ;
+    int n = b.length() ;
+    
+    int lcs[m + 1][n + 1];  
+ 
+    for (int i = 0; i <= m; i++)  
+    {  
+        for (int j = 0; j <= n; j++)  
+        {  
+        if (i == 0 || j == 0)  
+            lcs[i][j] = 0;  
+      
+        else if (X[i - 1] == Y[j - 1])  
+            lcs[i][j] = lcs[i - 1][j - 1] + 1;  
+      
+        else
+            lcs[i][j] = max(lcs[i - 1][j], lcs[i][j - 1]);  
+        }  
+    }  
+          
+    /* lcs[m][n] contains length of LCS  
+    for X[0..n-1] and Y[0..m-1] */
+    return lcs[m][n];
+}
+
+int main(){
+    string str1,str2 ; 
+    cout << "Take two strings to find their LCS\n";
+    cin >> str1 >> str2 ; 
+    cout << "Length of LCS is " 
+         << LCS( str1 , str2 ); 
+
+    return 0 ;     
+}
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diff --git a/javascript/Binary_Search.js b/javascript/Binary_Search.js
new file mode 100644
index 0000000..2f4666b
--- /dev/null
+++ b/javascript/Binary_Search.js
@@ -0,0 +1,23 @@
+// Iterative function to implement Binary Search 
+let iterativeFunction = function (arr, x) { 
+   
+    let start=0, end=arr.length-1; 
+          
+    // Iterate while start not meets end 
+    while (start<=end){ 
+  
+        // Find the mid index 
+        let mid=Math.floor((start + end)/2); 
+   
+        // If element is present at mid, return True 
+        if (arr[mid]===x) return true; 
+  
+        // Else look in left or right half accordingly 
+        else if (arr[mid] < x)  
+             start = mid + 1; 
+        else
+             end = mid - 1; 
+    } 
+   
+    return false; 
+}
\ No newline at end of file