From dbc69aae3380845d3483116820a89a5f9b625ac9 Mon Sep 17 00:00:00 2001
From: Vinay Mallela <69084915+MD571@users.noreply.github.com>
Date: Sat, 28 Oct 2023 22:26:19 +0530
Subject: [PATCH] Create spiral_matrix.cpp

---
 C++/spiral_matrix.cpp | 73 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 73 insertions(+)
 create mode 100644 C++/spiral_matrix.cpp

diff --git a/C++/spiral_matrix.cpp b/C++/spiral_matrix.cpp
new file mode 100644
index 0000000..3727a62
--- /dev/null
+++ b/C++/spiral_matrix.cpp
@@ -0,0 +1,73 @@
+/Problem: Given an m x n matrix, return all elements of the matrix in spiral order.
+/*
+Input:
+n=3 m=3
+Matriz:
+2       3       5
+19      23      7
+17      13      11
+Output:
+vector => 2 3 5 7 11 13 17 19 23
+*/
+#include <iostream> 
+#include<vector>
+#include<cmath>
+using namespace std;
+vector<int> SpiralOrder(vector<vector<int>> Mat,int n,int m) {
+	int val = 0,a[n][m];
+	int k = 0, l = 0; 
+	/*  k - starting row index
+        n - ending row index
+        l - starting column index
+        m - ending column index
+        i - iterator
+    */
+	vector<int> v;
+	//Spiral Form Order: (right->down->left->up->right->..->end)
+	while (k < n && l < m){
+        for (int i = l; i < m; ++i){//Right
+          v.push_back( Mat[k][i]);
+        }
+        k++;
+        for (int i = k; i < n; ++i){//Down
+            v.push_back(Mat[i][m-1]);
+        }
+        m--;
+        if (k < n) 
+        { 
+            for (int i = m-1; i >= l; --i){//left
+                v.push_back(Mat[n-1][i]);
+            }
+            n--; 
+        }
+        if (l < m) 
+        { 
+            for (int i = n-1; i >= k; --i){//up
+               v.push_back( Mat[i][l]);
+            }
+            l++; 
+        }
+	}
+	return v;
+}
+int main() 
+{ 
+	int m,n,i,j=0;
+    cin>>n>>m;
+    vector<vector<int>> Mat;
+    for(i=0;i<n;i++){
+		vector<int>a;
+		for(j=0;j<m;j++){
+			int v;
+            cin>>v;
+			a.push_back(v);
+        }
+		Mat.push_back(a);
+    }
+	vector<int> ans=SpiralOrder(Mat,n,m);
+	for(auto x:ans){
+		cout<<x<<" ";
+	}cout<<endl;
+	return 0; 
+} 
+//Time Complexity: O(M*N). To traverse the matrix O(m*n) time is required.