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Policemen catch thieves

  • Difficulty Level : Medium
  • Last Updated : 03 Sep, 2021
Geek Week

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.

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')
            pol.add(i);
            else if (arr[i] == 'T')
            thi.add(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 (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 */

Python3




# Python3 program to find maximum
# number of thieves caught
 
# Returns maximum number of thieves
# that can be caught.
def policeThief(arr, n, k):
    i = 0
    l = 0
    r = 0
    res = 0
    thi = []
    pol = []
 
    # store indices in list
    while i < n:
        if arr[i] == 'P':
            pol.append(i)
        elif arr[i] == 'T':
            thi.append(i)
        i += 1
 
    # track lowest current indices of
    # thief: thi[l], police: pol[r]
    while l < len(thi) and r < len(pol):
         
        # can be caught
        if (abs( thi[l] - pol[r] ) <= k):
            res += 1
            l += 1
            r += 1
             
        # increment the minimum index
        elif thi[l] < pol[r]:
            l += 1
        else:
            r += 1
 
    return res
 
# Driver program
if __name__=='__main__':
    arr1 = ['P', 'T', 'T', 'P', 'T']
    k = 2
    n = len(arr1)
    print(("Maximum thieves caught: {}".
         format(policeThief(arr1, n, k))))
 
    arr2 = ['T', 'T', 'P', 'P', 'T', 'P']
    k = 2
    n = len(arr2)
    print(("Maximum thieves caught: {}".
          format(policeThief(arr2, n, k))))
 
    arr3 = ['P', 'T', 'P', 'T', 'T', 'P']
    k = 3
    n = len(arr3)
    print(("Maximum thieves caught: {}".
          format(policeThief(arr3, n, k))))
 
# This code is contributed by `jahid_nadim`
Output
Maximum thieves caught: 2
Maximum thieves caught: 3
Maximum thieves caught: 3

Time Complexity: O(N) 
Auxiliary Space: O(N)

Following method works in O(1) space complexity



Approach:

This approach takes the following steps:

  1. First find the left most police and thief and store the indices. There can be two cases:
  2. CASE 1: If the distance between the police and thief <= k (given), the thief can be caught, so increment the res counter
  3. CASE 2: If the distance between the police and thief >= k, the current thief cannot be caught by the current police
    1. For CASE 2, if the police is behind the thief, we need to find the next police and check if it can catch the current thief
    2. if the thief is behind the police, we need to find the next thief and check if the current police can catch the thief
  4. Repeat the process until we find the next police and thief pair, and increment result counter if conditions are met, i,e, CASE 1.

Algorithm:
1. Initialize the current lowest indices of policeman in pol and thief in thi variable as -1.
2 Find the lowest index of policeman and thief.
3 If lowest index of either policeman or thief remain -1 then return 0.
4 If |pol – thi| <=k then make an allotment and find the next policeman and thief.
5 Else increment the min(pol , thi) to the next policeman or thief found.
6 Repeat the above two steps until we can find the next policeman and thief.
7 Return the number of allotments made.
Below is the implementation of the above algorithm.

Java




// Java program to find maximum number of
// thieves caught
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Returns maximum number of thieves that can
    // be caught.
    static int policeThief(char arr[], int n, int k)
    {
 
        int pol = -1, thi = -1, res = 0;
 
        // store the first index of police in pol
        for (int i = 0; i < n; i++) {
            if (arr[i] == 'P') {
                pol = i;
                break;
            }
        }
 
        // store the first index of thief in thi
        for (int i = 0; i < n; i++) {
            if (arr[i] == 'T') {
                thi = i;
                break;
            }
        }
 
        // return 0 if no police OR no thief found
        if (thi == -1 || pol == -1)
            return 0;
 
        // loop to increase res iff distance between
        // police and thief <= k
        while (pol < n && thi < n) {
 
            // thief can be caught
            if (Math.abs(pol - thi) <= k) {
 
                pol++;
 
                // to find the index of next police for next
                // iteration
                while (pol < n && arr[pol] != 'P') {
                    pol++;
                }
 
                // to find the index of next thief for next
                // iteration
                thi = thi + 1;
                while (thi < n && arr[thi] != 'T') {
                    thi++;
                    ;
                }
 
                // increment res, as the thief has been
                // caugh
                res++;
            }
 
            // thief cannot be caught as dist > k
            else if (thi < pol) {
 
                // as index of thief is behind police,
                // we need to find the next thief and check
                // if it can be caught by the current police
                //(it will be checked in the next iteration)
                // Hence, find the index of next thief
                thi++;
                while (thi < n && arr[thi] != 'T') {
                    thi++;
                }
            }
            else {
 
                // as the index of police is behind the
                // thief, it cannot catch the thief. Hence,
                // we need the index of next police and
                // check if it can catch the current thief
                //(it will be checked in the next iteration)
                pol++;
                while (pol < n && arr[pol] != 'P') {
                    pol++;
                }
            }
        }
 
        return res;
    }
 
    // Driver code starts
    public static void main(String[] args)
    {
 
        char arr1[] = { 'P', 'T', 'T', 'P', 'T' };
        int n = arr1.length;
        int k = 2;
        System.out.println("Maximum thieves caught: "
                           + policeThief(arr1, n, k));
 
        char arr2[] = { 'T', 'T', 'P', 'P', 'T', 'P' };
        n = arr2.length;
        k = 2;
        System.out.println("Maximum thieves caught: "
                           + policeThief(arr2, n, k));
 
        char arr3[] = { 'P', 'T', 'P', 'T', 'T', 'P' };
        n = arr3.length;
        k = 3;
        System.out.println("Maximum thieves caught: "
                           + policeThief(arr3, n, k));
    }
}
 
// Driver code ends

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 policeThief(char arr[], int n, int k)
{
    // Initialize the current lowest indices of
    // policeman in pol and thief in thi variable as -1
    int pol = -1, thi = -1, res = 0;
 
    // Find the lowest index of policemen
    for (int i = 0; i < n; i++) {
        if (arr[i] == 'P') {
            pol = i;
            break;
        }
    }
 
    // Find the lowest index of thief
    for (int i = 0; i < n; i++) {
        if (arr[i] == 'T') {
            thi = i;
            break;
        }
    }
 
    // If lowest index of either policemen or thief remain
    // -1 then return 0
    if (thi == -1 || pol == -1)
        return 0;
 
    while (pol < n && thi < n) {
 
        // can be caught
        if (abs(pol - thi) <= k) {
 
            pol = pol + 1;
            while (pol < n && arr[pol] != 'P') {
                pol = pol + 1;
            }
 
            thi = thi + 1;
            while (thi < n && arr[thi] != 'T') {
                thi = thi + 1;
            }
 
            res++;
        }
 
        // increment the current min(pol , thi) to
        // the next policeman or thief found
        else if (thi < pol) {
            thi = thi + 1;
            while (thi < n && arr[thi] != 'T') {
                thi = thi + 1;
            }
        }
        else {
            pol = pol + 1;
            while (pol < n && arr[pol] != 'P') {
                pol = pol + 1;
            }
        }
    }
 
    return res;
}
 
// Driver Code Starts.
 
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;
}
 
// Driver Code Ends
Output
Maximum thieves caught: 2
Maximum thieves caught: 3
Maximum thieves caught: 3

Time Complexity: O(N)
 Auxiliary Space: O(1)

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