diff --git a/problem1.cpp b/problem1.cpp
new file mode 100644
index 00000000..11ab8c3a
--- /dev/null
+++ b/problem1.cpp
@@ -0,0 +1,29 @@
+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Three line explanation of solution in plain english
+
+// Your code here along with comments explaining your approach
+// First, we go over the array from left to right to find the left running product
+// Then we go over the array from right to left, maintaining a right running product
+// The product of an array except a particular element will be the left product (existing result array) * right product
+
+class Solution {
+public:
+    vector<int> productExceptSelf(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> result(n);
+        result[0] = 1;
+        int leftProduct = 1; // left running product
+        for(int i=1; i<n; i++) {
+            leftProduct = leftProduct*nums[i-1];
+            result[i] = leftProduct;
+        }
+        int rightProduct = 1; // right running product
+        for(int i=n-2; i>=0; i--) {
+            rightProduct = rightProduct*nums[i+1];
+            result[i] = result[i]*rightProduct; 
+        }
+        return result;
+    }
+};
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diff --git a/problem2.cpp b/problem2.cpp
new file mode 100644
index 00000000..cc80b6a3
--- /dev/null
+++ b/problem2.cpp
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+// Time Complexity : O(m*n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Three line explanation of solution in plain english
+
+// Your code here along with comments explaining your approach
+// Start traversion from element at (0,0)
+// Collect the element at each step and put it in the answer vector
+// If we hit the boundary of the matrix (top most row, bottom most col, upper right element, bottom left element) change the direction
+
+class Solution {
+public:
+    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
+        int m = mat.size();
+        int n = mat[0].size();
+        vector<int> ans(m*n);
+        bool dir = true; // up
+        int r=0;
+        int c=0;
+        for(int i=0; i<m*n; i++) {
+            // collect the element
+            ans[i] = (mat[r][c]);
+            if(dir) { // up
+                if(r == 0 && c != n-1) {
+                    c++;
+                    dir = false; // flip direction when we hit boundary
+                } else if(c == n-1) {
+                    r++;
+                    dir = false; // flip direction when we hit boundary
+                } else {
+                    r--;
+                    c++;
+                }
+            } else { // down
+                if(c == 0 && r != m-1) {
+                    r++;
+                    dir = true; // flip direction when we hit boundary
+                } else if(r == m-1) {
+                    c++;
+                    dir = true; // flip direction when we hit boundary
+                } else {
+                    r++;
+                    c--;
+                }
+            }
+        }
+        return ans;
+    }
+};
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diff --git a/problem3.cpp b/problem3.cpp
new file mode 100644
index 00000000..89fdf469
--- /dev/null
+++ b/problem3.cpp
@@ -0,0 +1,47 @@
+// Time Complexity : O(m*n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Three line explanation of solution in plain english
+
+// Your code here along with comments explaining your approach
+// Keep track of the top, bottom, left, right boundaries of the matrix
+// Once we consume a row or column, shrink the appropriate boundaries
+// Check the base case (bounds check) inside the while loop as well, as we are mutating the bounds inside the loop
+
+class Solution {
+public:
+    vector<int> spiralOrder(vector<vector<int>>& matrix) {
+        int m = matrix.size(); int n = matrix[0].size();
+        int top = 0; int bottom = m - 1;
+        int left = 0; int right = n - 1;
+        vector<int> result;
+        while(left <= right && top <= bottom) {
+            for(int j=left; j<=right; j++) {
+                result.push_back(matrix[top][j]);
+            }
+            top++;
+
+            if(left <= right && top <= bottom) {
+                for(int i=top; i<=bottom; i++) {
+                    result.push_back(matrix[i][right]);
+                }
+            }
+            right--;
+
+            if(left <= right && top <= bottom) {
+                for(int j=right; j>=left; j--) {
+                    result.push_back(matrix[bottom][j]);
+                }
+            }
+            bottom--;
+
+            if(left <= right && top <= bottom) {
+                for(int i=bottom; i>=top; i--) {
+                    result.push_back(matrix[i][left]);
+                }
+            }
+            left++;
+        }
+        return result;
+    }
+};
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