24 - Amazon Top Interview Questions


Problem Statement :


You are given a list of four integers nums, each between 0 and 9, in a fixed order. By placing the operators +, -, *, and / (where / is integer division) between the numbers, and grouping them with parentheses, determine whether it is possible to get the value 24.

Constraints

n = 4 where n is the length of nums

0 ≤ nums[i] ≤ 9

Example 1

Input

nums = [5, 2, 7, 8]

Output

True

Explanation

We can do (5 * 2 - 7) * 8 = 24

Example 2

Input

nums = [7, 9, 7, 4]

Output

True

Explanation

We can make 24 with (7 - (9 / 7)) * 4. Note that / is integer division so the result of 9 / 7 is floored.


Solution :



title-img 




                        Solution in C++ :

vector<int> count(vector<int>& nums, int l, int r) {
    if (l == r) return {nums[l]};
    vector<int> ret;
    for (int j = l; j < r; j++) {
        vector<int> left = count(nums, l, j);
        vector<int> right = count(nums, j + 1, r);
        for (auto& n : left) {
            for (auto& p : right) {
                ret.push_back(n + p);
                ret.push_back(n * p);
                ret.push_back(n - p);
                if (p != 0) ret.push_back(n / p);
            }
        }
    }
    return ret;
}
bool solve(vector<int>& nums) {
    int l = 0, r = 3;
    auto dos = count(nums, l, r);
    int ret = 0;
    for (auto& n : dos) ret += n == 24;
    return ret > 0;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    char[] ops = {'+', '-', '*', '/'};

    public boolean solve(int[] nums) {
        return check(nums);
    }

    private boolean check(int[] arr) {
        if (arr.length == 1) {
            return arr[0] == 24;
        }

        for (int i = 0; i < arr.length - 1; i++) {
            for (char op : ops) {
                int[] res = operate(arr, i, op);
                if (check(res)) {
                    return true;
                }
            }
        }

        return false;
    }

    private int[] operate(int[] orig, int idx, char op) {
        int[] res = new int[orig.length - 1];
        for (int i = 0; i < idx; i++) {
            res[i] = orig[i];
        }

        if (op == '+') {
            res[idx] = orig[idx] + orig[idx + 1];
        } else if (op == '-') {
            res[idx] = orig[idx] - orig[idx + 1];
        } else if (op == '*') {
            res[idx] = orig[idx] * orig[idx + 1];
        } else {
            res[idx] = orig[idx + 1] == 0 ? 0 : orig[idx] / orig[idx + 1];
        }

        for (int i = idx + 2; i < orig.length; i++) {
            res[i - 1] = orig[i];
        }

        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        self.res = False

        def go(i, ops, postfix):
            # no. of ops. used is never more than
            # index
            if i == 4 and ops == 3:
                if self.postorder_eval(postfix) == 24:
                    self.res = True
            if ops < i - 1:
                for op in "+-/*":
                    postfix.append(op)
                    go(i, ops + 1, postfix)
                    postfix.pop()

            if i < len(nums):
                postfix.append(nums[i])
                go(i + 1, ops, postfix)
                postfix.pop()

            return

        go(0, 0, [])
        return self.res

    # postfix is easy to evaluate and no brackets ;)
    def postorder_eval(self, postfix):
        res = []

        for c in postfix:
            if isinstance(c, str):
                b = int(res.pop())
                a = int(res.pop())
                if c == "+":
                    c = a + b
                elif c == "-":
                    c = a - b
                elif c == "*":
                    c = a * b
                elif c == "/":
                    if b == 0:
                        c = 0
                    else:
                        c = a // b
                res.append(c)
            else:
                res.append(c)

        return res[-1]


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