New Problem Statement

Police catch thieves/Police_catch_thieves.cpp

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+#include <iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+//returning maximum number of thieves that can be caught
+int policeThief(char arr[], int n, int k){
+   int caught = 0;
+   vector<int> thieves;
+   vector<int> policemen;
+   for (int i = 0; i < n; i++) {  // here we are stroing indices in the vectors
+      if (arr[i] == 'P')
+         policemen.push_back(i);
+      else if (arr[i] == 'T')
+         thieves.push_back(i);
+   }
+   int thief = 0, police = 0; // tracking the lower indices
+   while (thief < thieves.size() && police < policemen.size()) {
+      if (abs(thieves[thief] - policemen[police]) <= k) { // thieves that can be caught 
+         caught++; 
+         thief++;
+         police++;
+      }
+      else if (thieves[thief] < policemen[police])
+         thief++;
+      else
+         police++;
+   }
+   return caught;
+}
+int main(){
+   int k, n;
+   char arr2[] = {'P', 'T', 'T', 'P', 'P', 'T', 'T', 'T', 'T', 'P' };
+   k = 2;
+   n = sizeof(arr2) / sizeof(arr2[0]);
+   cout << "Maximum number of thieves that can be caught by police is :"<<policeThief(arr2, n, k);
+   return 0;
+}

Police catch thieves/Readme.md

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+# POLICEMEN CATCH THIEVES
+
+Given an array of size n that has the following specifications: 
+
+1-Each element in the array contains either a policeman or a thief.  
+
+2-Each policeman can catch only one thief.  
+
+3-A policeman cannot catch a thief who is more than K units away from the policeman. 
+
+We need to find the maximum number of thieves that can be caught.
+Examples: 
+## Running Tests
+
+### Input
+
+```bash
+  arr[] = {'P', 'T', 'T', 'P', 'T'},
+            k = 1.
+```
+### Output
+
+```bash 
+2
+```
+
+### Input
+```bash 
+arr[] = {'T', 'T', 'P', 'P', 'T', 'P'}, 
+            k = 2
+```
+
+### Output 
+```bash
+3
+```