Leetcode 1404. Number of Steps to Reduce a Number in Binary Representation to One

Problem Explanation

The problem is based on manipulating the binary representation of a given number, s until we arrive at the number one (1). The rules for manipulation are simple and include:

Rule 1: If the current number is even, it must be divided by 2.

Rule 2: If the current number is odd, add 1 to it.

Our task is to calculate and return the number of steps taken to achieve this.

Approach to Solution:

The given problem can be modelled into an approach where we continuously check the last digit or the least significant bit (LSB) and accordingly apply the rules until we reach binary representation of 1.

1. If the least significant bit is '0', it means the number represented is even. As per rule, such number should be divided by 2. In binary representation, dividing by 2 is equivalent to removing the last digit. So, we simply remove the last digit and increment our steps counter.

2. If the least significant bit is '1' and it's not the only digit (number is not 1), it means the number represented is odd. As per rule, such number should be incremented by 1. In binary representation, incrementing an odd number by 1 leads to a carry propagating through all the 1's in LSB to the next 0's. So, we simulate this operation, increment our steps counter by 1 to account for the increment operation and then increment it for each 1 and corresponding 0 encountered.

3. Finally, if the binary number is '1', we can simply return the current steps counter as no more steps are needed.

To illustrate this, take the binary number '1101' as example.

• On the first iteration, the last digit (or LSB) is '1', its an odd number. We increment 1 and carry propagates through 1's to next 0. Updated binary after increment operation is '1110' and steps counter becomes 4 (1 for increment, 1 for each '1' and 2 for '0').
• On second iteration, the LSB is '0', its an even number. We divide it by 2 by removing LSB. Updated binary is '111' and steps becomes 5.
• On third iteration, the LSB is '1', it's an odd number. We increment 1 and all 1's turn to 0 and we are left with '1000'. Steps becomes 9.
• On fourth, fifth and sixth iteration, the LSB is '0', its an even number, we divide it by 2 by removing LSB and we are left with '1'. Count becomes 6.
• On seventh iteration, binary is '1' and we stop here and return the count of steps : '6'

Python Solution

1
2python 
3class Solution:
4    def numSteps(self, s: str) -> int:
5        steps = 0
6        while s != '1':
7            if s[-1] == '0':
8                s = s[:-1]  # removing last character or dividing by 2 
9                steps += 1
10            else:
11                carry = 0
12                s_new = ''
13                for ch in reversed(s):  
14                    # If current bit is 1, flip it to 0 and propagate carry
15                    if ch == '1':
16                        s_new = '0' + s_new
17                        carry = 1
18                    else:
19                        # If current bit is 0 and there is a carry, flip it to 1 and clear carry
20                        if carry == 1:
21                            s_new = '1' + s_new
22                            carry = 0
23                        else:
24                            s_new = '0' + s_new
25                # if there is still carry left, add it to front
26                if carry == 1:
27                    s_new = '1' + s_new
28                s = s_new
29                steps += 1
30        return steps

Java Solution

1
2java 
3class Solution {
4    public int numSteps(String s) {
5        int steps = 0;
6        while (!s.equals("1")) {
7            if (s.charAt(s.length()-1) == '0') {
8                s = s.substring(0, s.length()-1);  // removing last character or dividing by 2 
9                steps += 1;
10            } else {
11                int carry = 0;
12                String s_new = "";
13                for (int i = s.length()-1; i >= 0; --i) {  
14                    // If current bit is 1, flip it to 0 and propagate carry
15                    if (s.charAt(i) == '1') {
16                        s_new = '0' + s_new;
17                        carry = 1;
18                    } else {
19                        // If current bit is 0 and there is a carry, flip it to 1 and clear carry
20                        if (carry == 1) {
21                            s_new = '1' + s_new;
22                            carry = 0;
23                        } else {
24                            s_new = '0' + s_new;
25                        }
26                    } 
27                }
28                // if there is still carry left, add it to front
29                if (carry == 1) {
30                    s_new = '1' + s_new;
31                }
32                s = s_new;
33                steps += 1;
34            }
35        }
36        return steps;
37    }
38}

JavaScript Solution

1
2javascript
3class Solution {
4    numSteps(s) {
5        let steps = 0;
6        while (s !== '1') {
7            if (s.charAt(s.length-1) === '0') {
8                s = s.substring(0, s.length-1);  // removing last character or dividing by 2 
9                steps += 1;
10            } else {
11                let carry = 0;
12                let s_new = "";
13                for (let i = s.length-1; i >= 0; --i) {  
14                    // If current bit is 1, flip it to 0 and propagate carry
15                    if (s.charAt(i) === '1') {
16                        s_new = '0' + s_new;
17                        carry = 1;
18                    } else {
19                        // If current bit is 0 and there is a carry, flip it to 1 and clear carry
20                        if (carry === 1) {
21                            s_new = '1' + s_new;
22                            carry = 0;
23                        } else {
24                            s_new = '0' + s_new;
25                        }
26                    } 
27                }
28                // if there is still carry left, add it to front
29                if (carry === 1) {
30                    s_new = '1' + s_new;
31                }
32                s = s_new;
33                steps += 1;
34            }
35        }
36        return steps;
37    }
38}

C++ Solution

1
2cpp 
3class Solution {
4public:
5    int numSteps(string s) {
6        int steps = 0;
7        while (s != "1") {
8            if (s.back() == '0') {
9                s.pop_back();  // removing last character or dividing by 2 
10                steps += 1;
11            } else {
12                int carry = 0;
13                string s_new = "";
14                for (int i = s.length()-1; i >= 0; --i) {  
15                    // If current bit is 1, flip it to 0 and propagate carry
16                    if (s[i] == '1') {
17                        s_new = '0' + s_new;
18                        carry = 1;
19                    } else {
20                        // If current bit is 0 and there is a carry, flip it to 1 and clear carry
21                        if (carry == 1) {
22                            s_new = '1' + s_new;
23                            carry = 0;
24                        } else {
25                            s_new = '0' + s_new;
26                        }
27                    } 
28                }
29                // if there is still carry left, add it to front
30                if (carry == 1) {
31                    s_new = '1' + s_new;
32                }
33                s = s_new;
34                steps += 1;
35            }
36        }
37        return steps;
38    }
39};

C# Solution

1
2csharp 
3using System;
4using System.Linq;
5
6public class Solution {
7    public static int NumSteps(string s) {
8        int steps = 0;
9        while (s != "1") {
10            if (s[s.Length - 1] == '0') {
11                s = s.Remove(s.Length - 1);  // removing last character or dividing by 2 
12                steps++;
13            } else {
14                int carry = 0;
15                string s_new = String.Empty;
16                for (int i = s.Length - 1; i >= 0; i--) {  
17                    // If current bit is 1, flip it to 0 and propagate carry
18                    if (s[i] == '1') {
19                        s_new = '0' + s_new;
20                        carry = 1;
21                    } else {
22                        // If current bit is 0 and there is a carry, flip it to 1 and clear carry
23                        if (carry == 1) {
24                            s_new = '1' + s_new;
25                            carry = 0;
26                        } else {
27                            s_new = '0' + s_new;
28                        }
29                    }
30                }
31                // if there is still carry left, add it to front
32                if (carry == 1) {
33                    s_new = '1' + s_new;
34                }
35                s = s_new;
36                steps++;
37            }
38        }
39        return steps;
40    }
41}

Conclusions:

The problem explores the knowledge of binary arithmetic and string manipulations. The solutions are intuitive and work in O(N) time complexity with the python, java and C# solutions having a higher space complexity due to string concatenation operations. In JavaScript and C++ solutions the string concatenation operation overhead is eliminated as they do not use the '+' operator to append to string and instead they use the substring and pop_back functions respectively reducing the space complexity.

String concatenation using the '+' operator has a time complexity of O(n) which can increase execution time significantly for large inputs as new memory allocation and copying of characters is involved. However, with languages like C++ and JavaScript providing alternatives like pop_back and substring, the space overhead can be greatly reduced allowing the program to perform well even for large inputs.

In simpler terms, the key idea for this problem is to use bitwise operations to decide the next action based on whether the number is odd or even till we reach 1. Having a solid understanding of bitwise operations and how they correlate to simple arithmetic operations like division provides a basic yet effective solution to the problem.

All four solutions involve processing each bit in the binary representation and making decisions based on its value. While this is essential, how the binary representation is stored and manipulated can impact both run-time efficiency and memory usage. Therefore, when implementing such algorithms, careful consideration should be given to how best to store and manipulate data to ensure optimal performance.

Lastly, the implementation highlights that while all four languages - Python, Java, JavaScript, and C# - provide similar procedural control structures, data types and operations, the more nuanced aspects of how they handle string and memory management can lead to subtle differences in implementation and performance of the solution.