964

964. Least Operators to Express Number (Hard)

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /). For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

• The division operator (/) returns rational numbers.
• There are no parentheses placed anywhere.
• We use the usual order of operations: multiplication and division happen before addition and subtraction.
• It is not allowed to use the unary negation operator (-). For example, "x - x" is a valid expression as it only uses subtraction, but "-x + x" is not because it uses negation.

We would like to write an expression with the least number of operators such that the expression equals the given target. Return the least number of operators used.

 

Example 1:

Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3.
The expression contains 5 operations.

Example 2:

Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.
The expression contains 8 operations.

Example 3:

Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100.
The expression contains 3 operations.

 

Constraints:

• 2 <= x <= 100
• 1 <= target <= 2 * 108

Companies:
Bloomberg

Related Topics:
Math, Dynamic Programming

Solution 1.

// OJ: https://leetcode.com/problems/least-operators-to-express-number/
// Author: github.com/lzl124631x
// Time: O(?)
// Space: O(?)
// Ref: https://leetcode.com/problems/least-operators-to-express-number/discuss/208445/c%2B%2B-recursive-easy-to-understand
class Solution {
public:
    int leastOpsExpressTarget(int x, int target) {
        if (x > target) return min(target * 2 - 1, (x - target) * 2); // either adding x/x target times or subtract x/x (x - target) times from x
        if (x == target) return 0;
        long long prod = x, times = 0;
        while (prod < target) { // greedily multiply x as much as possible
            times++;
            prod *= x;
        }
        if (prod == target) return times; // target is power `times` of x
        int subtract = INT_MAX;
        if (prod - target < target) subtract = times + 1 + leastOpsExpressTarget(x, prod - target); // Unproved heuristic: If `prod - target >= target`, it will take the same or more operations than what is required by `target`.
        int add = times + leastOpsExpressTarget(x, target - (prod / x)); // using addition
        return min(subtract, add);
    }
};