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:
/) returns rational numbers.-). 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 <= 1001 <= target <= 2 * 108This approach involves recursively breaking down the target number into sums and differences of powers of the base number x. We utilize memoization to store the results of subproblems to avoid redundant calculations.
This Python code recursively finds the minimum number of operations to express the target using powers of x. By storing intermediate results in a memoization table, it improves efficiency significantly.
Time Complexity: O(logxtarget) due to recursive calls and memoization.
Space Complexity: O(logxtarget) for the memoization table.
This approach uses dynamic programming to build up a solution iteratively. It constructs an array to store the minimal operators required for each subproblem up to the target.
This Java solution utilizes a dynamic programming map to store intermediate results and handle overlapping subproblems efficiently.
Python
JavaScript
Time Complexity: O(logxtarget) due to memoization of subproblems.
Space Complexity: O(logxtarget) for the storage of intermediate results.
This method uses backtracking with a greedy approach. The key is to choose the nearest power of x to either over or under target and recursively assess the number of operations needed.
This C++ solution utilizes an unordered_map to cache results. Similar to DP, it recursively exploits the nearest multiples to minimize the number of operators using power and modulo calculations.
Java
Time Complexity: O(logxtarget) - logarithmic exploration based on division.
Space Complexity: O(logxtarget) for caching states.
| Approach | Complexity |
|---|---|
| Recursive Approach with Memoization | Time Complexity: O(logxtarget) due to recursive calls and memoization. |
| Dynamic Programming Approach | Time Complexity: O(logxtarget) due to memoization of subproblems. |
| Greedy Backtracking Approach | Time Complexity: O(logxtarget) - logarithmic exploration based on division. |
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