What is the time complexity of Least Operators to Express Number?

The optimal solution has a time complexity of O(log(target)).

What is the space complexity of Least Operators to Express Number?

The space complexity is O(log(target)).

What data structures are used to solve Least Operators to Express Number?

This problem uses Math, Dynamic Programming, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Least Operators to Express Number?

Insight 1: The problem can be solved using dynamic programming (memoization) due to overlapping subproblems in the recursive exploration of possible expressions. Insight 2: The core idea is to consider two options: either adding or subtracting multiples of x^p to reach the target. We can express the target as q * x + r, where q is the quotient and r is the remainder when dividing the target by x. Then, we recursively solve for q and q+1, and choose the option that leads to fewer operators.

Which companies ask LeetCode 964. Least Operators to Express Number in interviews?

This problem is frequently asked at Snap.

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Least Operators to Express Number - LeetCode 964 Solution

Least Operators to Express Number - Complete Solution Guide

Least Operators to Express Number is LeetCode problem 964, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to find the minimum number of operators (+, -, *, /) needed to express a given 'target' number using only a single positive integer 'x'. We can create expressions like 'x * x + x / x - x', and we need to find the expression that equals 'target' using the fewest operators. The division operator returns rational numbers, there are no parentheses, standard order of operations applies, and unary negation is not allowed. We are given constraints on x and target.

Solution Approach

The solution uses a top-down dynamic programming approach with memoization to explore the space of possible expressions. The `solve` function recursively calculates the minimum number of operators needed to reach a specific value `t` using powers of `x` starting from `x^p`. It considers expressing `t` as `q * x + r`. It calculates the optimal number of operations needed using `q * x^p + r * x^(p-1)` or `(q+1) * x^p - (x-r) * x^(p-1)`, where q is target/x and r is target%x. The function explores both adding and subtracting multiples of x^p to get closer to the target. The `memo` dictionary stores the results of previously computed subproblems to avoid redundant calculations.

Step-by-Step Algorithm

Step 1: Initialize a memoization table (dictionary in Python, HashMap in Java, unordered_map in C++, custom hash table in C) to store the results of previously computed subproblems. The key is a tuple (p, t), where p is the power of x, and t is the target value.
Step 2: Define a recursive function `solve(p, t)` that takes the power `p` of x and the target value `t` as input. This function returns the minimum number of operators needed to express `t` using powers of `x` starting from `x^p`.
Step 3: In the base cases for `solve(p, t)`:
- If `t == 0`, return 0 (no more operators needed).
- If `t == 1`, return `cost(p)`, which is 2 if p is 0 and p if p is greater than 0
- If `p > 35`, return a large value (infinity) to prune the search space. Powers of x beyond 35 are likely to be too large to be useful. In C/C++, INT_MAX is used, while Python uses float('inf').
Step 4: Calculate the quotient `q` and remainder `r` when dividing `t` by `x`: `q = t // x` and `r = t % x`.
Step 5: Consider two options:
- Option 1: Express `t` as `q * x + r`. The number of operators needed is `solve(p + 1, q) + r * cost(p)`. This represents using r instances of x^p. The cost for multiplying by r is r * (cost of one x^p)
- Option 2: Express `t` as `(q + 1) * x - (x - r)`. The number of operators needed is `solve(p + 1, q + 1) + (x - r) * cost(p)`. This means you round up and subtract, for which you will need another number of operations.
Step 6: Take the minimum of the two options: `min(solve(p + 1, q) + r * cost(p), solve(p + 1, q + 1) + (x - r) * cost(p))`. This step ensures that we pick the better expression.
Step 7: Store the result in the memoization table: `memo[(p, t)] = result`.
Step 8: Return the result from the `solve` function.
Step 9: The main function `leastOpsExpressTarget(x, target)` calls `solve(0, target)` and subtracts 1 from the result. This final subtraction removes the artificial operator count that is caused by getting to 1 or 0

Key Insights

Insight 1: The problem can be solved using dynamic programming (memoization) due to overlapping subproblems in the recursive exploration of possible expressions.
Insight 2: The core idea is to consider two options: either adding or subtracting multiples of x^p to reach the target. We can express the target as q * x + r, where q is the quotient and r is the remainder when dividing the target by x. Then, we recursively solve for q and q+1, and choose the option that leads to fewer operators.
Insight 3: We need to consider the 'cost' of using x^p, which is 'p' operations (multiplications) to build x^p from x. Special case p=0, the cost is 2 (division operation is needed to get 1).
Insight 4: The problem implicitly requires considering powers of 'x', up to a point where adding or subtracting x^p becomes more expensive than the reduction in the value. The constraint `p > 35` serves as a pruning condition for the recursion, as x <= 100, and continuing to calculate higher powers of x leads to diminishing returns (and avoids potential integer overflow).

Complexity Analysis

Time Complexity: O(log(target))

Space Complexity: O(log(target))

Topics

This problem involves: Math, Dynamic Programming, Memoization.

Companies

Asked at: Snap.

Least Operators to Express Number | LeetCode 964

Hard

Math Dynamic Programming Memoization

Time: O(log(target))Space: O(log(target))

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int leastOpsExpressTarget(int x, int target) {
        Map<String, Integer> memo = new HashMap<>();

        // Helper function to generate key for memoization
        // (Java doesn't allow tuple keys directly)
        // Can be improved by using a custom Pair class if needed
        
        // Helper function to calculate cost
        int cost(int p) {
            if (p == 0) {
                return 2;
            }
            return p;
        }

        // Recursive function to solve the problem
        int solve(int p, int t) {
            String key = p + "," + t;
            if (memo.containsKey(key)) {
                return memo.get(key);
            }

            if (t == 0) {
                return 0;
            }
            if (t == 1) {
                return cost(p);
            }

            if (p > 35) {
                return Integer.MAX_VALUE;
            }

            int q = t / x;
            int r = t % x;

            int res;
            if (r == 0) {
                res = solve(p + 1, q);
                memo.put(key, res);
                return res;
            }

            int op1 = solve(p + 1, q) + r * cost(p);
            int op2 = solve(p + 1, q + 1) + (x - r) * cost(p);

            res = Math.min(op1, op2);
            memo.put(key, res);
            return res;
        }

        return solve(0, target) - 1;
    }
}

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Problem Info

Description

Problem 964: Least Operators to Express Number - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Memoization

Companies

Snap

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:964

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

14 steps

1
Step 1: Initialize a memoization table (dictionary in Python, HashMap in Java, unordered_map in C++, custom hash table in C) to store the results of previously computed subproblems. The key is a tuple (p, t), where p is the power of x, and t is the target value.
2
Step 2: Define a recursive function `solve(p, t)` that takes the power `p` of x and the target value `t` as input. This function returns the minimum number of operators needed to express `t` using powers of `x` starting from `x^p`.
3
Step 3: In the base cases for `solve(p, t)`:
4
- If `t == 0`, return 0 (no more operators needed).
5
- If `t == 1`, return `cost(p)`, which is 2 if p is 0 and p if p is greater than 0
6
- If `p > 35`, return a large value (infinity) to prune the search space. Powers of x beyond 35 are likely to be too large to be useful. In C/C++, INT_MAX is used, while Python uses float('inf').
7
Step 4: Calculate the quotient `q` and remainder `r` when dividing `t` by `x`: `q = t // x` and `r = t % x`.
8
Step 5: Consider two options:
9
- Option 1: Express `t` as `q * x + r`. The number of operators needed is `solve(p + 1, q) + r * cost(p)`. This represents using r instances of x^p. The cost for multiplying by r is r * (cost of one x^p)
10
- Option 2: Express `t` as `(q + 1) * x - (x - r)`. The number of operators needed is `solve(p + 1, q + 1) + (x - r) * cost(p)`. This means you round up and subtract, for which you will need another number of operations.
11
Step 6: Take the minimum of the two options: `min(solve(p + 1, q) + r * cost(p), solve(p + 1, q + 1) + (x - r) * cost(p))`. This step ensures that we pick the better expression.
12
Step 7: Store the result in the memoization table: `memo[(p, t)] = result`.
13
Step 8: Return the result from the `solve` function.
14
Step 9: The main function `leastOpsExpressTarget(x, target)` calls `solve(0, target)` and subtracts 1 from the result. This final subtraction removes the artificial operator count that is caused by getting to 1 or 0

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(target))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(log(target))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is logarithmic because the target value is reduced by a factor of x in each recursive call. The number of levels in the recursion tree is roughly log_x(target). At each level, we perform a constant amount of work to calculate the quotient and remainder. The space complexity is also logarithmic because the size of the memoization table is proportional to the number of distinct (p, t) pairs visited during the recursion, which is related to the depth of the recursion.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bottom-up Dynamic Programming

Instead of using recursion with memoization, we can use a bottom-up dynamic programming approach to build the solution iteratively. We create a table dp[i][j] where dp[i][j] stores the minimum number of operators needed to reach value j using powers of x up to x^i. The table is built starting from the base cases and iterating to larger values.

Time: O(log(target) * target)Space: O(log(target) * target)

Trade-offs: The bottom-up approach can be more space-efficient in some cases if we only need the final result, but might be less intuitive to implement than the top-down approach.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming (memoization) due to overlapping subproblems in the recursive exploration of possible expressions.
2Insight 2: The core idea is to consider two options: either adding or subtracting multiples of x^p to reach the target. We can express the target as q * x + r, where q is the quotient and r is the remainder when dividing the target by x. Then, we recursively solve for q and q+1, and choose the option that leads to fewer operators.
3Insight 3: We need to consider the 'cost' of using x^p, which is 'p' operations (multiplications) to build x^p from x. Special case p=0, the cost is 2 (division operation is needed to get 1).
4Insight 4: The problem implicitly requires considering powers of 'x', up to a point where adding or subtracting x^p becomes more expensive than the reduction in the value. The constraint `p > 35` serves as a pruning condition for the recursion, as x <= 100, and continuing to calculate higher powers of x leads to diminishing returns (and avoids potential integer overflow).

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: target = 1. The algorithm should correctly return the cost of getting to 1 from x^0, which is 2.
Edge case 2: target is a large number close to INT_MAX. The algorithm avoids integer overflow by using appropriate data types (long long in C++, int in Python) and/or pruning the search space.
Edge case 3: x = 2, target = even/odd number. The algorithm correctly considers adding and subtracting powers of x to efficiently express the target.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization, resulting in exponential time complexity due to redundant calculations.

✗Common mistake 2: Incorrectly calculating the cost of using x^p (the number of multiplication operations needed to obtain x^p). For p=0, the cost is division of x/x to get to 1, hence 2 operations are used

✗Common mistake 3: Integer overflow when calculating large values of t or cost. Use long long (C++) or handle overflow explicitly.

✗Common mistake 4: Forgetting to subtract 1 at the end when using the recursive function.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Compiler optimization. Compilers can use similar dynamic programming techniques to optimize arithmetic expressions, minimizing the number of operations needed to compute a given result.

🏦Real-world scenario 2: Financial modeling. Financial models often involve complex calculations, and similar optimization techniques can be used to reduce the computational cost of these models.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Least Operators to Express Number"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?