HackerRank Minimum Operations problem solution

In this HackerRank Minimum Operations problem solution, There are n boxes in front of you. For each i, box i contains r[i] red balls, g[i] green balls, and b[i] blue balls.

You want to separate the balls by their color. In each operation, you can pick a single ball from some box and put it into another box. The balls are separated if no box contains balls of more than one color.

Debug the given function min_operations and compute the minimal number of operations required to separate the balls.

Note: In this problem you can modify at most six lines of code and you cannot add any new lines.

HackerRank Minimum Operations problem solution in C++.

#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <algorithm> 
#include <vector> 
#include <cmath> 
#include <iostream> 
#include <map> 
using namespace std; 

int dp[110][1<<3];

int min_operations(vector <int> red, vector <int> green, vector <int> blue) {

    int n = (int)red.size(), i, j;
    for (i = 0; i <= n; i++) {
        for (j = 0; j < 8; j++) {
            dp[i][j] = 1<<30;
        }
    }

    dp[0][0] = 0;
    for (i = 0; i < n; i++){
        for (j = 0; j < 8; j++){
            dp[i + 1][j | 1] = min(dp[i + 1][j | 1], dp[i][j] + green[i] + blue[i]);
            dp[i + 1][j | 2] = min(dp[i + 1][j | 2], dp[i][j] + red[i] + blue[i]);
            dp[i + 1][j | 4] = min(dp[i + 1][j | 4], dp[i][j] + red[i] + green[i]);
        }
    }
    j = 0;
    for (i = 0; i < n; i++){
        if (green[i]) j |= 2;
        if (red[i]) j |= 1;
        if (blue[i]) j |= 4;
    }

    if (dp[n][j] >= (1<<30))
        dp[n][j] = -1;

    return dp[n][j];
}

int main() {

    int n, r, g, b;
    cin >> n;
    vector<int> red, blue, green;

    for(int i = 0; i < n; i++){

        cin >> r >> g >> b;
        red.push_back(r);
        green.push_back(g);
        blue.push_back(b);
    }

    cout << min_operations(red, green, blue) << "n";
    return 0;
}

Minimum Operations problem solution in Java

import java.util.*;

class MinimumOperations {
    private static final Scanner scan = new Scanner(System.in);
    int n, r ,g, b;
    int[][] dp = new int[110][1<<3];

    Vector<Integer> red = new Vector();
    Vector<Integer> green = new Vector();
    Vector<Integer> blue = new Vector();

    public void get() {
         n = scan.nextInt();

        for (int i = 0; i < n; i++) {
            r = scan.nextInt();
            g = scan.nextInt();
            b = scan.nextInt();
            red.add(r);
            green.add(g);
            blue.add(b);
        }
    }

    public void minOperations() {
        int i, j;
        for (i = 0; i <= n; i++) {
            for (j = 0; j < 8; j++) {
                dp[i][j] = (1<<30);
            }
        }
        dp[0][0] = 0;
        for (i = 0; i < n; i++){
            for (j = 0; j < 8; j++){
                dp[i + 1][j | 1] = Math.min(dp[i + 1][j | 1], dp[i][j] + green.get(i) + blue.get(i));
                dp[i + 1][j | 2] = Math.min(dp[i + 1][j | 2], dp[i][j] + red.get(i) + blue.get(i));
                dp[i + 1][j | 4] = Math.min(dp[i + 1][j | 4], dp[i][j] + red.get(i) + green.get(i));
            }
        }
        j = 0;
        for (i = 0; i < n; i++){
            if (green.get(i) != 0) j |= 2;
            if (red.get(i) != 0) j |= 1;
            if (blue.get(i) != 0) j |= 4;
        }
        if (dp[n][j] >= (1<<30)) dp[n][j] = -1;
            System.out.println(dp[n][j]);
    }
}
class Solution {
    public static void main(String[] args) {
        MinimumOperations obj = new MinimumOperations();
        obj.get();
        obj.minOperations();
    }
}