Policemen catch thieves

Last Updated : 27 Mar, 2025

Given an array arr, where each element represents either a policeman (P) or a thief (T). The objective is to determine the maximum number of thieves that can be caught under the following conditions:

Each policeman (P) can catch only one thief (T).
A policeman can only catch a thief if the distance between them is at most k units.

Your task is to find the maximum number of thieves that can be caught following these rules.

Examples:

Input: arr[] = ['P', 'T', 'T', 'P', 'T'], k = 1
Output: 2.
Explanations: Here maximum 2 thieves can be caught, first
policeman catches first thief and second police man can catch either second or third thief.

Input: arr[] = ['T', 'T', 'P'], k = 1
Output: 1
Explanations: Here second thief can be caught by the police.

Table of Content

[Naive Approach] Using Nested loops - O(n*k) Time and O(n) Space

The main idea is that for every policeman, we search for a thief within a range of 2*k + 1 in a sequential manner. If a thief is found, we insert their index into the caught array to ensure that the same thief is not caught multiple times.

Step by Step implementation:

We traverse the array from left to right.
If the current element is a policeman (P), search for a thief (T) within a range of [i - k, i + k].
Each thief can be caught only once, so we track caught thieves using an auxiliary array.
If a thief is found within this range who is not already caught, mark them as caught and increase the count.
Return the count of caught thieves.

C++

// C++ program to find maximum number of thieves // caught #include <bits/stdc++.h> using namespace std;  // Returns maximum number of thieves that can // be caught. int catchThieves(vector<char> &arr, int k) {     int n = arr.size();          // To mark if a thief is caught     vector<bool> caught(n, false);           // Stores the number of caught thieves     int count = 0;      for (int i = 0; i < n; ++i) {         if (arr[i] == 'P') {              int start = max(0, i - k);              int end = min(n - 1, i + k);                           // Search for an uncaught thief within range             for (int j = start; j <= end; ++j) {                 if (arr[j] == 'T' && !caught[j]) {                      caught[j] = true;                      count++;                      break;                  }             }         }     }     return count;  } int main() {     int k = 1;     vector<char> arr = { 'P', 'T', 'T', 'P', 'T' };     cout<< catchThieves(arr, k) << endl; }

Java

// Java program to find maximum number of thieves // caught import java.util.*;  public class GfG {          // Function to find the maximum number of thieves     // that can be caught      static int catchThieves(char[] arr, int k) {         int n = arr.length;                  // Boolean array to keep track of caught thieves         boolean[] caught = new boolean[n];                   // Count of caught thieves         int count = 0;                   // Traverse the array to find policemen         for (int i = 0; i < n; i++) {                          // Found a policeman             if (arr[i] == 'P') {                  int start = Math.max(0, i - k);                  int end = Math.min(n - 1, i + k);                                   // Search for an uncaught thief within range                 for (int j = start; j <= end; j++) {                     if (arr[j] == 'T' && !caught[j]) {                          caught[j] = true;                          count++;                          break;                      }                 }             }         }         return count;      }      public static void main(String[] args) {         int k = 1;         char[] arr = { 'P', 'T', 'T', 'P', 'T' };                  System.out.println(catchThieves(arr, k));      } }

Python

# Python program to find the maximum number of thieves caught  # Returns the maximum number of thieves that can be caught def catchThieves(arr, k):     n = len(arr)          # To mark if a thief is caught     caught = [False] * n            # Stores the number of caught thieves     count = 0            for i in range(n):         if arr[i] == 'P':               start = max(0, i - k)               end = min(n - 1, i + k)                            # Search for an uncaught thief within range             for j in range(start, end + 1):                 if arr[j] == 'T' and not caught[j]:                       caught[j] = True                       count += 1                       break            return count    if __name__ == "__main__":     k = 1     arr = ['P', 'T', 'T', 'P', 'T']     print(catchThieves(arr, k))

using System;  class GfG {          // Returns the maximum number of thieves that can be caught     static int catchThieves(char[] arr, int k) {         int n = arr.Length;                  // To mark if a thief is caught         bool[] caught = new bool[n];                   // Stores the number of caught thieves         int count = 0;                   for (int i = 0; i < n; i++)         {             if (arr[i] == 'P')             {                 int start = Math.Max(0, i - k);                 int end = Math.Min(n - 1, i + k);                                  // Search for an uncaught thief within range                 for (int j = start; j <= end; j++)                 {                     if (arr[j] == 'T' && !caught[j])                     {                         caught[j] = true;                         count++;                         break;                     }                 }             }         }         return count;     }      static void Main() {         int k = 1;         char[] arr = { 'P', 'T', 'T', 'P', 'T' };                  Console.WriteLine(catchThieves(arr, k));       } }

JavaScript

// JavaScript program to find the maximum number of thieves caught  // Returns the maximum number of thieves  // that can be caught function catchThieves(arr, k) {     let n = arr.length;          // To mark if a thief is caught     let caught = new Array(n).fill(false);           // Stores the number of caught thieves     let count = 0;           for (let i = 0; i < n; i++) {         if (arr[i] === 'P') {             let start = Math.max(0, i - k);             let end = Math.min(n - 1, i + k);                          // Search for an uncaught thief within range             for (let j = start; j <= end; j++) {                 if (arr[j] === 'T' && !caught[j]) {                      caught[j] = true;                     count++;                     break;                 }             }         }     }     return count; }  // Driver Code const k = 1; const arr = ['P', 'T', 'T', 'P', 'T'];  console.log(catchThieves(arr, k));

Output

[Expected Approach] Using Two Pointers - O(n) Time and O(1) Space

The main idea is to use two pointers (both begin from left side): one to track policemen and the other to track thieves.
We traverse the array while maintaining these pointers.
If a policeman and a thief are within the allowed range k, the thief is caught, and both pointers move forward.
If the thief is too far left, we move the thief pointer forward;
similarly, if the policeman is too far left, we move the policeman pointer forward.

Step by Step implementation:

Initialize Two Pointers: i for the next policeman ('P') and j for the next thief ('T'), along with a counter count = 0.
Traverse the Array: Move i to the next 'P' and j to the next 'T'.
Check Distance: If |i - j| ≤ k, catch the thief (count++), then move both pointers forward.
Adjust Pointers: If the thief (j) is too far left, move j forward; if the policeman (i) is too far left, move i forward.
Repeat Until End: Continue until either i or j reaches the end of the array. Return count.

C++

// C++ program to find maximum number of thieves // caught #include <bits/stdc++.h> using namespace std;  // Returns the maximum number of thieves // that can be caught using two pointers int catchThieves(vector<char> &arr, int k) {     int n = arr.size();          // Two pointers for policemen and thieves     int i = 0, j = 0;      int count = 0;      while (i < n && j < n) {                  // Move i to the next policeman         while (i < n && arr[i] != 'P') i++;          // Move j to the next thief         while (j < n && arr[j] != 'T') j++;          // If both policeman and thief exist and are within range k         if (i < n && j < n && abs(i - j) <= k) {                          // Catch the thief             count++;                           // Move to the next policeman             i++;                           // Move to the next thief             j++;          }                   // If the thief is too far left, move the thief pointer         else if (j < n && j < i) {             j++;         }                   // If the policeman is too far left,          // move the policeman pointer         else if (i < n && i < j) {             i++;         }     }     return count; }  int main() {     int k = 1;     vector<char> arr = { 'P', 'T', 'T', 'P', 'T' };     cout<< catchThieves(arr, k) << endl; }

Java

// Java program to find maximum number of thieves // caught import java.util.*;  public class GfG {          // Returns the maximum number of thieves that     // can be caught using two pointers     static int catchThieves(char[] arr, int k) {         int n = arr.length;          // Two pointers for policemen and thieves         int i = 0, j = 0;         int count = 0;          while (i < n && j < n) {                          // Move i to the next policeman             while (i < n && arr[i] != 'P') i++;              // Move j to the next thief             while (j < n && arr[j] != 'T') j++;              // If both policeman and thief exist              // and are within range k             if (i < n && j < n && Math.abs(i - j) <= k) {                                  // Catch the thief                 count++;                                    // Move to the next policeman                 i++;                                    // Move to the next thief                 j++;               }                           // If the thief is too far left,             // move the thief pointer             else if (j < n && j < i) {                 j++;             }                           // If the policeman is too far left,             // move the policeman pointer             else if (i < n && i < j) {                 i++;             }         }         return count;     }      public static void main(String[] args) {         int k = 1;         char[] arr = { 'P', 'T', 'T', 'P', 'T' };                  System.out.println(catchThieves(arr, k));      } }

Python

# Python program to find the maximum number of thieves caught  # Returns the maximum number of thieves  # that can be caught using two pointers def catchThieves(arr, k):     n = len(arr)          # Two pointers for policemen and thieves     i, j = 0, 0     count = 0      while i < n and j < n:                  # Move i to the next policeman         while i < n and arr[i] != 'P':             i += 1          # Move j to the next thief         while j < n and arr[j] != 'T':             j += 1          # If both policeman and thief exist         # and are within range k         if i < n and j < n and abs(i - j) <= k:                          # Catch the thief             count += 1                            # Move to the next policeman             i += 1                            # Move to the next thief             j += 1                    # If the thief is too far left,         # move the thief pointer         elif j < n and j < i:             j += 1                  # If the policeman is too far left,         # move the policeman pointer         elif i < n and i < j:             i += 1      return count   if __name__ == "__main__":     k = 1     arr = ['P', 'T', 'T', 'P', 'T']     print(catchThieves(arr, k))

using System;  class GfG {           static int catchThieves(char[] arr, int k) {         int n = arr.Length;                  // Two pointers for policemen and thieves         int i = 0, j = 0;          int count = 0;          while (i < n && j < n) {                          // Move i to the next policeman             while (i < n && arr[i] != 'P')                  i++;              // Move j to the next thief             while (j < n && arr[j] != 'T')                  j++;              // If both policeman and thief exist             // and are within range k             if (i < n && j < n && Math.Abs(i - j) <= k) {                                  // Catch the thief                 count++;                  // Move to the next policeman                 i++;                   // Move to the next thief                 j++;              }                          // If the thief is too far left, move the thief pointer             else if (j < n && j < i) {                 j++;             }                          // If the policeman is too far left,             // move the policeman pointer             else if (i < n && i < j) {                 i++;             }         }         return count;     }      static void Main() {         int k = 1;         char[] arr = { 'P', 'T', 'T', 'P', 'T' };                  Console.WriteLine(catchThieves(arr, k));       } }

JavaScript

// JavaScript program to find the maximum number of thieves caught  // Returns the maximum number of thieves  // that can be caught function catchThieves(arr, k) {     let n = arr.length;      // Two pointers for policemen and thieves     let i = 0, j = 0;     let count = 0;      while (i < n && j < n) {                  // Move i to the next policeman         while (i < n && arr[i] !== 'P') i++;          // Move j to the next thief         while (j < n && arr[j] !== 'T') j++;          // If both policeman and thief exist and are within range k         if (i < n && j < n && Math.abs(i - j) <= k) {                          // Catch the thief             count++;              // Move to the next policeman             i++;              // Move to the next thief             j++;         }                   // If the thief is too far left, move the thief pointer         else if (j < n && j < i) {             j++;         }                   // If the policeman is too far left, move the policeman pointer         else if (i < n && i < j) {             i++;         }     }     return count; }  // Driver Code const k = 1; const arr = ['P', 'T', 'T', 'P', 'T'];  console.log(catchThieves(arr, k));

Output

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Policemen catch thieves

[Naive Approach] Using Nested loops - O(n*k) Time and O(n) Space

[Expected Approach] Using Two Pointers - O(n) Time and O(1) Space

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