How do you solve Least Operators to Express Number on LeetCode?

Least Operators to Express Number (LeetCode #964) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding operator precedence is crucial for forming valid expressions.

What is the brute force solution for Least Operators to Express Number?

The brute force approach for Least Operators to Express Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of operations to reach the target. It explores all possible expressions formed by applying the operators repeatedly until the target is achieved. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Least Operators to Express Number?

Least Operators to Express Number uses the following concepts: Math, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Breadth-First Search.

What are the interview tips for Least Operators to Express Number?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process as you work through the problem to demonstrate your understanding. Consider both brute force and optimized approaches, as this shows depth in problem-solving.

What are common mistakes when solving Least Operators to Express Number?

Forgetting to handle division correctly, especially with integer division. Not considering negative results when using subtraction.

#964

Least Operators to Express Number

Hard

MathDynamic ProgrammingMemoizationDynamic ProgrammingBreadth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build up the minimum number of operations needed to reach each possible value up to the target. By storing previously computed results, we avoid redundant calculations.

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Algorithm

3 steps

1Step 1: Create a dictionary to store the minimum operations needed to reach each value.
2Step 2: Initialize the dictionary with the starting value x requiring 0 operations.
3Step 3: For each value in the dictionary, apply each operator, updating the dictionary with the minimum operations needed to reach new values.

solution.py13 lines

1def leastOperators(x, target):
2    dp = {x: 0}
3    for value in range(1, target + 1):
4        if value in dp:
5            ops = dp[value]
6            for op in ['+', '-', '*', '/']:
7                if op == '+': new_value = value + x
8                elif op == '-': new_value = value - x
9                elif op == '*': new_value = value * x
10                elif op == '/': new_value = value / x
11                if new_value <= target and (new_value not in dp or dp[new_value] > ops + 1):
12                    dp[new_value] = ops + 1
13    return dp.get(target, -1)

ℹ

Complexity note: The complexity is linear because we only iterate through values up to the target once, storing results in a dictionary to avoid redundant calculations.

1Understanding operator precedence is crucial for forming valid expressions.
2Dynamic programming can significantly reduce the number of operations by reusing previously computed results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.