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+Question:There are N people standing in a circle waiting to be executed. The counting out begins at some 
+point in the circle and proceeds around the circle in a fixed direction. In each step, a certain number 
+of people are skipped and the next person is executed. The elimination proceeds around the circle 
+(which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. 
+
+Given the total number of persons N and a number k which indicates that k-1 persons are skipped and the kth person is killed in a circle.
+The task is to choose the person in the initial circle that survives.
+
+
+Example:Input: N = 5 and k = 2
+Output: 3
+Explanation: Firstly, the person at position 2 is killed, 
+then the person at position 4 is killed, then the person at position 1 is killed. 
+Finally, the person at position 5 is killed. So the person at position 3 survives. 
+
+
+Input: N = 7 and k = 3
+Output: 4
+Explanations: The persons at positions 3, 6, 2, 7, 5, and 1 are killed in order, 
+and the person at position 4 survives.
+
+
+Code in Python Language:
+
+def Josh(person, k, index):
+
+  # Base case , when only one person is left
+  if len(person) == 1:
+    print(person[0])
+    return
+
+  # find the index of first person which will die
+  index = ((index+k)%len(person))
+
+   # remove the first person which is going to be killed
+  person.pop(index)
+
+  # recursive call for n-1 persons
+  Josh(person,k,index)
+
+# Driver Program to test above function
+n = 14 # specific n and k  values for original josephus problem
+k = 2
+k-=1   # (k-1)th person will be killed
+
+index = 0 
+
+# fill the person vector
+person=[]
+for i in range(1,n+1):
+  person.append(i)
+
+Josh(person,k,index)
+
+
+
+CODE IN C++
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+void Josh(vector<int> person, int k, int index)
+{
+    // Base case , when only one person is left
+    if (person.size() == 1) {
+        cout << person[0] << endl;
+        return;
+    }
+
+    // find the index of first person which will die
+    index = ((index + k) % person.size());
+
+    // remove the first person which is going to be killed
+    person.erase(person.begin() + index);
+
+    // recursive call for n-1 persons
+    Josh(person, k, index);
+}
+
+int main()
+{
+    int n = 14; // specific n and k  values for original
+                // josephus problem
+    int k = 2;
+    k--; // (k-1)th person will be killed
+    int index
+        = 0; // The index where the person which will die
+
+    vector<int> person;
+    // fill the person vector
+    for (int i = 1; i <= n; i++) {
+        person.push_back(i);
+    }
+
+    Josh(person, k, index);
+}