What is the normal way to debug nested for loop? - c++

Question

I am writing code for Leetcode problem 38. Count and Say. It doesn't pass the cases, so I add some cout to debug. Please tell me is there a normal way to debug nested for loop, where should I add the cout expression. I don't want to know how to modify the code to pass the cases.
Here is my code:
class Solution {
public:
string countAndSay(int n) {
string cur("1");
while (--n) {
string tmp = cur;
string next;
for (int i = 0; i < tmp.size();) {
cout << "i:" << i << endl;
int count = 1;
for (int j = i + 1; j < tmp.size(); j++) {
if (tmp[j] != tmp[0]) {
break;
}
count++;
}
cout << "count:" << count << endl;
next += std::to_string(count) + tmp[0];
cout << "cur:" << cur << endl;
i += count;
}
cur = next;
cout << n << cur << endl;
}
return cur;
}
};

Answer 1

You're gonna have to use a debugger for that, and step by step go through your algorithm to find the bugs. It's hard to debug someone else's algorithm.
This'll pass through:
#include <string>
struct Solution {
static const std::string countAndSay(int n) {
if (not n) {
return "";
}
std::string res = "1";
while (--n) {
std::string curr = "";
for (int index = 0; index < res.size(); index++) {
int count = 1;
while ((index + 1 < res.size()) and (res[index] == res[index + 1])) {
count++;
index++;
}
curr += std::to_string(count) + res[index];
}
res = curr;
}
return res;
}
};
Java Solutions
class Solution {
public String countAndSay(int n) {
if (n == 1)
return "1";
String prev = countAndSay(n - 1);
StringBuilder str = new StringBuilder();
int i = 0;
while (i < prev.length()) {
char curr = prev.charAt(i);
int j = 0;
while (i + j < prev.length() && prev.charAt(i + j) == curr)
j++;
str.append(j);
str.append(curr);
i += j;
}
return str.toString();
}
}
Here is one of LeetCode's solutions using regex:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class Solution {
public String countAndSay(int n) {
String currSeq = "1";
// Pattern to match the repetitive digits
String regexPattern = "(.)\\1*";
Pattern pattern = Pattern.compile(regexPattern);
for (int i = 1; i < n; ++i) {
Matcher m = pattern.matcher(currSeq);
StringBuffer nextSeq = new StringBuffer();
// each group contains identical and adjacent digits
while (m.find()) {
nextSeq.append(m.group().length() + String.valueOf(m.group().charAt(0)));
}
// prepare for the next iteration
currSeq = nextSeq.toString();
}
return currSeq;
}
}
and here's another LeetCode's solution also using Sliding Window:
class Solution {
public String countAndSay(int n) {
LinkedList<Integer> prevSeq = new LinkedList<Integer>();
prevSeq.add(1);
// Using -1 as the delimiter
prevSeq.add(-1);
List<Integer> finalSeq = this.nextSequence(n, prevSeq);
StringBuffer seqStr = new StringBuffer();
for (Integer digit : finalSeq) {
seqStr.append(String.valueOf(digit));
}
return seqStr.toString();
}
protected LinkedList<Integer> nextSequence(int n, LinkedList<Integer> prevSeq) {
if (n <= 1) {
// remove the delimiter before return
prevSeq.pollLast();
return prevSeq;
}
LinkedList<Integer> nextSeq = new LinkedList<Integer>();
Integer prevDigit = null;
Integer digitCnt = 0;
for (Integer digit : prevSeq) {
if (prevDigit == null) {
prevDigit = digit;
digitCnt += 1;
} else if (digit == prevDigit) {
// in the middle of the sub-sequence
digitCnt += 1;
} else {
// end of a sub-sequence
nextSeq.add(digitCnt);
nextSeq.add(prevDigit);
// reset for the next sub-sequence
prevDigit = digit;
digitCnt = 1;
}
}
// add the delimiter for the next recursion
nextSeq.add(-1);
return this.nextSequence(n - 1, nextSeq);
}
}
References
For additional details, you can see the Discussion Board. There are plenty of accepted solutions with a variety of languages and explanations, efficient algorithms, as well as asymptotic time/space complexity analysis1, 2 in there.