add codeforces solutions folder

leetcode/lc1582/spaceOptimized.cpp

@@ -0,0 +1,48 @@
+class Solution{
+    public:
+        int numSpecial(vector<vector<int>> &mat){
+            int ans = 0;
+            int m = mat.size();
+            int n = mat[0].size();
+            for(int row = 0;row<m;row++){
+                int colIndex = -1;
+                int count = 0;
+                for(int col = 0;col<n;col++){
+                    if(mat[row][col]==1){
+                        count++;
+                        colIndex = col;
+                    }
+                }
+                if(count==1){
+                bool good = true;
+            for(int r=0;r<m;r++){
+
+                if(r!=row && mat[r][colIndex]==1) {
+                    good = false;
+                    break;
+                }
+            }
+            if(good) ans++;
+            }
+            }
+            return ans;
+        }
+};
+
+/*
+Approach:
+
+We iterate through each row of the matrix and count how many 1s appear in that row.
+While scanning the row, we also record the column index where the 1 occurs. If a row
+contains exactly one 1, it becomes a candidate for a special position. For that candidate,
+we then check the entire column corresponding to that 1 to ensure no other row contains a 1
+in the same column. If the column also has exactly one 1 (the current position), then that
+position satisfies both conditions and we increment the answer. After checking all rows,
+the total number of such positions is returned.
+
+Time Complexity: O(mn + m²) in the worst case. We scan each row once (mn work), and for
+rows that contain exactly one 1 we scan the column (up to m operations). In practice this
+is usually closer to O(mn).
+
+Space Complexity: O(1) 
+*/