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+# Josephus Problem Algorithm (Recursive Solution)
+
+## Overview
+
+The **Josephus Problem** is a theoretical problem related to a group of people standing in a circle, where every `k`-th person is eliminated in each round until only one person remains. The challenge is to determine the position in which a person should stand to avoid elimination and be the last one standing.
+
+This repository contains a C program that solves the Josephus Problem using a recursive approach. The program takes two inputs:
+- `n`: the total number of people in the circle.
+- `k`: the step count at which people are eliminated.
+
+## Theory
+
+In the Josephus Problem:
+- If there is only one person, they will be the survivor by default (base case).
+- For any other number of people, the problem can be reduced recursively. Each time a person is eliminated, the circle effectively shrinks, and the position of the next person changes.
+
+The recursive formula for finding the safe position is:
+\[ \text{Josephus}(n, k) = (\text{Josephus}(n-1, k) + k) \% n \]
+where:
+- `n` is the number of people remaining,
+- `k` is the step count.
+
+Using this recursive relation, we start with one person (`n = 1`) and build up to the solution for `n` people.
+
+## Code Example
+
+The main code logic is implemented in the function `josephus(int n, int k)`. Here’s the pseudocode outline:
+1. Base case: If there’s only one person, they are the survivor.
+2. Recursive case: For more than one person, recursively calculate the safe position for `n - 1` people, adjust by the step count `k`, and use modulo `n` to wrap around the circle.
+
+### Time Complexity
+
+The time complexity of this recursive approach is **O(n)**. This is because each recursive call reduces the problem size by 1, and there are `n` calls in total (one for each person).
+
+### Space Complexity
+
+The space complexity is **O(n)** due to the recursive stack, which holds up to `n` frames as the function calls itself for each person until reaching the base case.
+
+## Usage
+
+### Input
+The program prompts the user for:
+1. The number of people in the circle (`n`).
+2. The step count for elimination (`k`).
+
+### Output
+The program outputs the safe position, which is the position a person should occupy to be the last one standing.
+
+### Sample Output
+
+For `n = 7` people and `k = 3` as the step count:
+```plaintext
+Enter the number of people (n): 7
+Enter the step count (k): 3
+The safe position is 4
+```
+
+### How to Run
+1. Compile the program using a C compiler:
+   ```bash
+   gcc josephus.c -o josephus
+   ```
+2. Run the compiled program:
+   ```bash
+   ./josephus
+   ```
+
+## Conclusion
+
+The Josephus Problem is a classic example of a recursive solution that reduces the problem size with each step. The algorithm demonstrates the power of recursion in solving combinatorial problems efficiently by using a mathematical recurrence relation. This solution, while optimal in time complexity, has a space complexity of `O(n)` due to recursion depth, making it efficient for moderate values of `n`.
diff --git a/Recursion/ Josephus Problem Algorithm/program.c b/Recursion/ Josephus Problem Algorithm/program.c
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+#include <stdio.h>
+
+// Recursive function to solve the Josephus Problem
+int josephus(int n, int k) {
+    // Base case: if there's only one person, they are the survivor
+    if (n == 1)
+        return 0;
+    // Recurrence relation to find the safe position
+    return (josephus(n - 1, k) + k) % n;
+}
+
+int main() {
+    int n, k;
+
+    // Get the number of people and the step count from the user
+    printf("Enter the number of people (n): ");
+    scanf("%d", &n);
+    printf("Enter the step count (k): ");
+    scanf("%d", &k);
+
+    // Find the safe position (adjusted by +1 to match 1-based indexing)
+    int position = josephus(n, k) + 1;
+    printf("The safe position is %d\n", position);
+    
+    return 0;
+}