Spirally traversing a matrix

Author: sachuverma

DSA Crack Sheet/README.md

@@ -43,6 +43,7 @@
 
 ### Matrix
 
+- [Spirally traversing a matrix](https://practice.geeksforgeeks.org/problems/spirally-traversing-a-matrix-1587115621/1# "view question") - [Cpp Solution](./solutions/Spirally%20traversing%20a%20matrix.cpp)  
 - []( "view question") - [Cpp Solution](./solutions/.cpp)  
 
 ### Strings

DSA Crack Sheet/solutions/Spirally traversing a matrix.cpp

@@ -0,0 +1,82 @@
+/*
+Spirally traversing a matrix
+============================
+
+Given a matrix of size R*C. Traverse the matrix in spiral form.
+
+Example 1:
+Input:
+R = 4, C = 4
+matrix[][] = {{1, 2, 3, 4},
+           {5, 6, 7, 8},
+           {9, 10, 11, 12},
+           {13, 14, 15,16}}
+Output: 
+1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10
+
+Example 2:
+Input:
+R = 3, C = 4  
+matrix[][] = {{1, 2, 3, 4},
+           {5, 6, 7, 8},
+           {9, 10, 11, 12}}
+Output: 
+1 2 3 4 8 12 11 10 9 5 6 7
+
+Your Task:
+You dont need to read input or print anything. Complete the function spirallyTraverse() that takes matrix, R and C as input parameters and returns a list of integers denoting the spiral traversal of matrix. 
+
+Expected Time Complexity: O(R*C)
+Expected Auxiliary Space: O(R*C)
+
+Constraints:
+1 <= R, C <= 100
+0 <= matrixi <= 100
+*/
+
+vector<int> spirallyTraverse(vector<vector<int>> matrix, int r, int c)
+{
+  int count = r * c;
+  vector<int> ans;
+  int tl = 0, tr = c - 1, lt = 0, lb = r - 1;
+  while (count > 0)
+  {
+
+    for (int i = tl; i <= tr && count > 0; ++i)
+    {
+      ans.push_back(matrix[lt][i]);
+      count--;
+    }
+    lt++;
+    if (count <= 0)
+      break;
+
+    for (int i = lt; i <= lb && count > 0; ++i)
+    {
+      ans.push_back(matrix[i][tr]);
+      count--;
+    }
+    tr--;
+    if (count <= 0)
+      break;
+
+    for (int i = tr; i >= tl && count > 0; --i)
+    {
+      ans.push_back(matrix[lb][i]);
+      count--;
+    }
+    lb--;
+    if (count <= 0)
+      break;
+
+    for (int i = lb; i >= lt && count > 0; --i)
+    {
+      ans.push_back(matrix[i][tl]);
+      count--;
+    }
+    tl++;
+    if (count <= 0)
+      break;
+  }
+  return ans;
+}