Introduction
In this article, we will explore efficient solution in C, C++, Java, and Python to crack the ZigZag Conversion LeetCode question.
Cracking LeetCode questions is a valuable skill for every programmer. One such intriguing problem is the ZigZag Conversion question.
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By understanding the problem, implementing optimized solutions, and leveraging the power of these programming languages, we will master this coding challenge.
Understanding the ZigZag Conversion Problem
Before diving into the solutions, let’s understand the ZigZag Conversion problem. The problem statement is as follows:
“You are given a string s and an integer numRows representing the number of rows in a ZigZag pattern. Your task is to convert the string into a ZigZag pattern by reading the characters row by row and return the converted string.”
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For example, let’s say we have the string “PAYPALISHIRING” and numRows is 3. The ZigZag conversion would look like this:
P A H N
A P L S I I G
Y I RThe resulting converted string would be “PAHNAPLSIIGYIR”.
Now that we have a clear understanding of the problem, let’s dive into efficient solutions for cracking the ZigZag Conversion LeetCode question using different programming languages.
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Cracking the ZigZag Conversion in C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* convert(char* s, int numRows) {
if (numRows == 1 || numRows >= strlen(s)) {
return s;
}
char** zigzag = (char**)malloc(numRows * sizeof(char*));
for (int i = 0; i < numRows; i++) {
zigzag[i] = (char*)malloc((strlen(s) + 1) * sizeof(char));
memset(zigzag[i], 0, (strlen(s) + 1) * sizeof(char));
}
int row = 0, step = 1;
for (int i = 0; i < strlen(s); i++) {
zigzag[row][strlen(zigzag[row])] = s[i];
if (row == 0) {
step = 1;
} else if (row == numRows - 1) {
step = -1;
}
row += step;
}
char* result = (char*)malloc((strlen(s) + 1) * sizeof(char));
memset(result, 0, (strlen(s) + 1) * sizeof(char));
for (int i = 0; i < numRows; i++) {
strcat(result, zigzag[i]);
free(zigzag[i]);
}
free(zigzag);
return result;
}
int main() {
char* s = "PAYPALISHIRING";
int numRows = 3;
char* result = convert(s, numRows);
printf("Converted String: %s\n", result);
free(result);
return 0;
}Cracking the ZigZag Conversion in C++
#include <iostream>
#include <string>
std::string convert(std::string s, int numRows) {
if (numRows == 1 || numRows >= s.length()) {
return s;
}
std::vector<std::string> zigzag(numRows, "");
int row = 0, step = 1;
for (char c : s) {
zigzag[row] += c;
if (row == 0) {
step = 1;
} else if (row == numRows - 1) {
step = -1;
}
row += step;
}
std::string result;
for (std::string str : zigzag) {
result += str;
}
return result;
}
int main() {
std::string s = "PAYPALISHIRING";
int numRows = 3;
std::string result = convert(s, numRows);
std::cout << "Converted String: " << result << std::endl;
return 0;
}Cracking the ZigZag Conversion in Java
public class ZigZagConversion {
public String convert(String s, int numRows) {
if (numRows == 1 || numRows >= s.length()) {
return s;
}
StringBuilder[] zigzag = new StringBuilder[numRows];
for (int i = 0; i < numRows; i++) {
zigzag[i] = new StringBuilder();
}
int row = 0, step = 1;
for (char c : s.toCharArray()) {
zigzag[row].append(c);
if (row == 0) {
step = 1;
} else if (row == numRows - 1) {
step = -1;
}
row += step;
}
StringBuilder result = new StringBuilder();
for (StringBuilder sb : zigzag) {
result.append(sb);
}
return result.toString();
}
public static void main(String[] args) {
String s = "PAYPALISHIRING";
int numRows = 3;
ZigZagConversion converter = new ZigZagConversion();
String result = converter.convert(s, numRows);
System.out.println("Converted String: " + result);
}
}Cracking the ZigZag Conversion in Python
def convert(s: str, numRows: int) -> str:
if numRows == 1 or numRows >= len(s):
return s
zigzag = [''] * numRows
row, step = 0, 1
for c in s:
zigzag[row] += c
if row == 0:
step = 1
elif row == numRows - 1:
step = -1
row += step
return ''.join(zigzag)
s = "PAYPALISHIRING"
numRows = 3
result = convert(s, numRows)
print("Converted String:", result)Also Read: Longest Substring Without Repeating Characters
FAQs
The ZigZag Conversion LeetCode question is a problem where you are given a string and an integer representing the number of rows in a ZigZag pattern. The task is to convert the string into a ZigZag pattern by reading the characters row by row and return the converted string.
The efficient solutions for cracking the ZigZag Conversion LeetCode question involve iterating through the string and placing the characters in the appropriate rows based on the ZigZag pattern. We can achieve this using extra space or perform the conversion in-place without using any extra space.
This article provides efficient solutions for the ZigZag Conversion LeetCode question in C, C++, Java, and Python.
Yes, you can solve the ZigZag Conversion LeetCode question using languages other than C, C++, Java, and Python. The logic and approach remain the same; only the syntax may vary.
Yes, the solutions provided in this article for cracking the ZigZag Conversion LeetCode question are optimized and efficient. They are designed to solve the problem with the best time and space complexity.
You can practice the ZigZag Conversion LeetCode question on the LeetCode platform. LeetCode offers a wide range of coding problems and algorithmic challenges to help you enhance your programming skills.
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Conclusion
In this article, we explored efficient solution to crack the ZigZag Conversion LeetCode question using the programming languages of C, C++, Java, and Python.
We discussed the problem statement, understood the requirements, and implemented optimized solutions to solve the problem.
By leveraging the power of algorithms and efficient coding techniques, we can tackle even the most challenging coding problems. So, keep practicing, keep coding, and keep cracking those LeetCode questions!
