# 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:

```Input : arr[] = {'P', 'T', 'T', 'P', 'T'},
k = 1.
Output : 2.
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', 'P', 'T', 'P'},
k = 2.
Output : 3.

Input : arr[] = {'P', 'T', 'P', 'T', 'T', 'P'},
k = 3.
Output : 3.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A brute force approach would be to check all feasible sets of combinations of police and thief and return the maximum size set among them. Its time complexity is exponential and it can be optimized if we observe an important property.

An efficient solution is to use a greedy algorithm. But which greedy property
to use can be tricky. We can try using: “For each policeman from the left catch the nearest possible thief.” This works for example three given above but fails for example two as it outputs 2 which is incorrect.
We may also try: “For each policeman from the left catch the farthest possible thief”. This works for example two given above but fails for example three as it outputs 2 which is incorrect. A symmetric argument can be applied to show that traversing for these from the right side of the array also fails. We can observe that thinking irrespective of the
policeman and focusing on just the allotment works:

1. Get the lowest index of policeman p and thief t. Make an allotment
if |p-t| <= k and increment to the next p and t found.
2. Otherwise increment min(p, t) to the next p or t found.
3. Repeat above two steps until next p and t are found.
4. Return the number of allotments made.

Below is the implementation of the above algorithm. It uses vectors to
store the indices of police and thief in the array and processes them.

## C++

```// C++ program to find maximum number of thieves
// caught
#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

// Returns maximum number of thieves that can
// be caught.
int policeThief(char arr[], int n, int k)
{
int res = 0;
vector<int> thi;
vector<int> pol;

// store indices in the vector
for (int i = 0; i < n; i++) {
if (arr[i] == 'P')
pol.push_back(i);
else if (arr[i] == 'T')
thi.push_back(i);
}

// track lowest current indices of
// thief: thi[l], police: pol[r]
int l = 0, r = 0;
while (l < thi.size() && r < pol.size()) {

// can be caught
if (abs(thi[l] - pol[r]) <= k) {
res++;
l++;
r++;
}

// increment the minimum index
else if (thi[l] < pol[r])
l++;
else
r++;
}

return res;
}

// Driver program
int main()
{
int k, n;

char arr1[] = { 'P', 'T', 'T', 'P', 'T' };
k = 2;
n = sizeof(arr1) / sizeof(arr1[0]);
cout << "Maximum thieves caught: "
<< policeThief(arr1, n, k) << endl;

char arr2[] = { 'T', 'T', 'P', 'P', 'T', 'P' };
k = 2;
n = sizeof(arr2) / sizeof(arr2[0]);
cout << "Maximum thieves caught: "
<< policeThief(arr2, n, k) << endl;

char arr3[] = { 'P', 'T', 'P', 'T', 'T', 'P' };
k = 3;
n = sizeof(arr3) / sizeof(arr3[0]);
cout << "Maximum thieves caught: "
<< policeThief(arr3, n, k) << endl;

return 0;
}
```

## Java

```// Java program to find maximum number of
// thieves caught
import java.util.*;
import java.io.*;

class GFG
{
// Returns maximum number of thieves
// that can be caught.
static int policeThief(char arr[], int n, int k)
{
int res = 0;
ArrayList<Integer> thi = new ArrayList<Integer>();
ArrayList<Integer> pol = new ArrayList<Integer>();

// store indices in the ArrayList
for (int i = 0; i < n; i++) {
if (arr[i] == 'P')
else if (arr[i] == 'T')
}

// track lowest current indices of
// thief: thi[l], police: pol[r]
int l = 0, r = 0;
while (l < thi.size() && r < pol.size()) {

// can be caught
if (Math.abs(thi.get(l) - pol.get(r)) <= k) {
res++;
l++;
r++;
}

// increment the minimum index
else if (thi.get(l) < pol.get(r))
l++;
else
r++;
}
return res;
}

// Driver program
public static void main(String args[])
{
int k, n;
char arr1[] =new char[] { 'P', 'T', 'T',
'P', 'T' };
k = 2;
n = arr1.length;
System.out.println("Maximum thieves caught: "
+policeThief(arr1, n, k));

char arr2[] =new char[] { 'T', 'T', 'P', 'P',
'T', 'P' };
k = 2;
n = arr2.length;
System.out.println("Maximum thieves caught: "
+policeThief(arr2, n, k));

char arr3[] = new char[]{ 'P', 'T', 'P', 'T',
'T', 'P' };
k = 3;
n = arr3.length;
System.out.println("Maximum thieves caught: "
+policeThief(arr3, n, k));
}
}

/* This code is contributed by Danish kaleem */
```

Output:

```Maximum thieves caught: 2
Maximum thieves caught: 3
Maximum thieves caught: 3
```

The time complexity of the above approach in O(N) where N is the size of the array.

This article is contributed by Satish Srinivas. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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