completed problem 1 & problem 2

problem1.cpp

@@ -0,0 +1,60 @@
+// Time Complexity : O(n!)
+// Space Complexity : O(n^2)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+
+// Your code here along with comments explaining your approach
+// Start with row 1 and go upto nth row
+// For each cell, try placing a queen after checking there are no queens in the same col or the diagonals
+// Backtrack after placing the queen to check other valid configurations
+
+class Solution {
+private:
+    bool isValid(int row, int col, int n, vector<string>& currState) {
+        // check column
+        for(int i=row-1; i>=0; i--) {
+            if(currState[i][col] == 'Q') return false;
+        }
+
+        // check diag - up - right
+        int i = row-1; int j = col+1;
+        while(i >=0 && j < n) {
+            if(currState[i][j] == 'Q') return false;
+            i--;
+            j++;
+        }
+        // check diag - up - left
+        i = row-1; j = col-1;
+        while(i >= 0 && j >= 0) {
+            if(currState[i][j] == 'Q') return false;
+            i--;
+            j--;
+        }
+
+        return true;
+    }
+    void helper(int row, int n, vector<string>& currState, vector<vector<string>>& valid) {
+        if(row == n) {
+            valid.push_back(currState);
+            return;
+        }
+        string currRow(n, '.');
+        for(int i=0; i<n; i++){
+            if(isValid(row, i, n, currState)) {
+                currRow[i]='Q';
+                currState.push_back(currRow);
+                helper(row+1, n, currState, valid);
+                currRow[i] = '.';
+                currState.pop_back();
+            }
+        }
+    }
+public:
+    vector<vector<string>> solveNQueens(int n) {
+        vector<string> currState;
+        vector<vector<string>> valid;
+        helper(0, n, currState, valid);
+        return valid;
+    }
+};

problem2.cpp

@@ -0,0 +1,47 @@
+// Time Complexity : O(m*n*4^L), L = length of word
+// Space Complexity : O(L)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+
+// Your code here along with comments explaining your approach
+// Start DFS from cells which have the starting char of the word
+// Explore valid cells in all 4 directions, use a # symbol to mark a cell as visited
+// Backtrack # to unmark as visited once that path has been explored
+
+class Solution {
+private:
+    vector<vector<int>> dirs = {{0,1},{0,-1},{1,0},{-1,0}};
+    void dfs(int r, int c, int m, int n, vector<vector<char>>& board, int idx, string word, bool& flag) {
+        if(idx == word.length()) {
+            flag = true;
+            return;
+        }
+        if(flag) return;
+        for(auto& dir : dirs) {
+            int nr = r + dir[0]; int nc = c + dir[1];
+            if(nr >= 0 && nr < m && nc >= 0 && nc < n && board[nr][nc] == word[idx] && board[nr][nc] != '#') {
+                board[nr][nc] = '#';
+                dfs(nr, nc, m, n, board, idx+1, word, flag);
+                board[nr][nc] = word[idx];
+            }
+        }
+    }
+public:
+    bool exist(vector<vector<char>>& board, string word) {
+        int m = board.size();
+        int n = board[0].size();
+        bool flag = false;
+        for(int i=0; i<m; i++) {
+            for(int j=0; j<n; j++) {
+                if(board[i][j] == word[0]) {
+                    char temp = board[i][j];
+                    board[i][j] = '#';
+                    dfs(i, j, m, n, board, 1, word, flag);
+                    board[i][j] = temp;
+                }
+            }
+        }
+        return flag;
+    }
+};