N queen problem

backtracking/n-queens/C++/nqueenProblem.cpp

@@ -0,0 +1,77 @@
+#include <bits/stdc++.h>
+#define N 4
+using namespace std;
+
+/* A utility function to print solution */
+void printSolution(int board[N][N])
+{
+    for (int i = 0; i < N; i++)
+    {
+        for (int j = 0; j < N; j++)
+            cout << " " << board[i][j] << " ";
+        printf("\n");
+    }
+}
+
+bool isSafe(int board[N][N], int row, int col)
+{
+    int i, j;
+    for (i = 0; i < col; i++)
+        if (board[row][i])
+            return false;
+
+    for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
+        if (board[i][j])
+            return false;
+
+    for (i = row, j = col; j >= 0 && i < N; i++, j--)
+        if (board[i][j])
+            return false;
+
+    return true;
+}
+
+bool solveNQUtil(int board[N][N], int col)
+{
+    if (col >= N)
+        return true;
+
+    for (int i = 0; i < N; i++)
+    {
+        if (isSafe(board, i, col))
+        {
+            board[i][col] = 1;
+
+            if (solveNQUtil(board, col + 1))
+                return true;
+
+            board[i][col] = 0; // BACKTRACK
+        }
+    }
+
+    return false;
+}
+
+bool solveNQ()
+{
+    int board[N][N] = {{0, 0, 0, 0},
+                       {0, 0, 0, 0},
+                       {0, 0, 0, 0},
+                       {0, 0, 0, 0}};
+
+    if (solveNQUtil(board, 0) == false)
+    {
+        cout << "Solution does not exist";
+        return false;
+    }
+
+    printSolution(board);
+    return true;
+}
+
+// driver program to test above function
+int main()
+{
+    solveNQ();
+    return 0;
+}

strings/Index of First Occurence/cpp/indexOfFirstOccurence.cpp

@@ -0,0 +1,26 @@
+// 28. Find the Index of the First Occurrence in a String
+
+class Solution
+{
+public:
+    int strStr(string haystack, string needle)
+    {
+        int m = haystack.size(), n = needle.size();
+        for (int i = 0; i <= m - n; i++)
+        {
+            int j = 0;
+            for (; j < n; j++)
+            {
+                if (haystack[i + j] != needle[j])
+                {
+                    break;
+                }
+            }
+            if (j == n)
+            {
+                return i;
+            }
+        }
+        return -1;
+    }
+};