# Do You Know How to Solve These Programming Problems?

Programming is about solving problems. In this piece, I’ve listed six programming problems from several sites that contain programming problems. The problems in this listed are sorted based on how difficult they are to solve — with number one on this list being the easiest, and number six being the most difficult to solve. Can you wrap your head around all of them?

I provided the solutions to these programming problems, coded in PHP, at the bottom of this article. You can choose a programming language of your choice to solve these problems.

Good luck solving these problems, and happy coding!

# Challenges

Let’s start with a relatively easy problem provided by HackerRank. This challenge is categorized as a warmup.

A challenge that’s considered easy, provided by LeetCode.

This problem is provided by Project Euler and is considered one of the more easy problems. It’s currently solved by more than 455,000 people.

This is the problem:

Another challenge from Project Euler. This one is a little bit harder than the previous problem. It’s solved by around 100,000 people.

If you’ve made it up to this point: Congratulations! It’s time to start the first hard challenge. This challenge is provided by Coderbyte.

Definitely the toughest challenge so far, this challenge is provided by LeetCode. Although it’s considered to be of medium difficulty, I found this one harder to solve than the Kaprekars Constant. This one requires you to know how linked lists work.

But let’s not go into too much detail — here’s the challenge:

# Solutions

A really great problem to get you started with a solution that’s really straightforward.

`<?phpfunction getFractionals(\$numbers) { \$length = count(\$numbers); \$results = [  'positive' => 0,  'negative' => 0,  'zero' => 0, ];  for (\$i = 0; \$i < \$length; \$i++) {  if (\$numbers[\$i] < 0) {   \$results['negative'] += 1;   } else if (\$numbers[\$i] > 0) {   \$results['positive'] += 1;   } else {   \$results['zero'] += 1;   } }  return [  \$results['positive'] / \$length,  \$results['negative'] / \$length,  \$results['zero'] / \$length   ];}print_r(getFractionals([1, 1, 0, -1, -1])); // [0.4, 0.4, 0.2]print_r(getFractionals([-4, 3, -9, 0, 4, 1])); // [0.5, 0.3333, 0.16667]`

Although this one’s a little bit harder than the first problem, you shouldn’t have too much trouble solving this one. I used a simple brute-force approach.

`<?phpfunction twoSum(\$numbers, \$target) {  for (\$i = 0; \$i < count(\$numbers); \$i++) {    for (\$j = \$i + 1; \$j < count(\$numbers); \$j++) {      if (\$numbers[\$j] + \$numbers[\$i] === \$target) {        return [\$i, \$j];      }    }  }}print_r(twoSum([2, 7, 11, 15], 9)); // [0, 1]print_r(twoSum([2, 7, 11, 15], 17)); // [0, 3]`

The solution I came up with has the advantage it can be used to find the biggest palindrome that’s the product of any two x-digit numbers.

I’ve also added stop conditions to avoid unnecessary extra looping.

`<?phpfunction isPalindrome(\$number) {  return (string) \$number === strrev((string) \$number);}function getBiggestPalindrome(\$digits) {  \$start = pow(10, \$digits) - 1;  \$max = 0;for (\$i = \$start; \$i > 0; \$i--) {    if (\$i * \$start <= \$max) {      break;      }for (\$j = \$start; \$j > 0; \$j--) {      \$product = \$i * \$j;if (\$product < \$max) {        break;        }if (\$product > \$max && isPalindrome(\$product)) {        \$max = \$product;      }    }  }return \$max;}echo getBiggestPalindrome(2); // 9009echo getBiggestPalindrome(3); // 906609, which is 993 * 913`

I solved the distinct powers problem by going the brute-force route.

Add each result to the array, and then remove duplicates from the array. The last step is to sort the array.

`<?phpfunction distinctPowers(\$min, \$max) { \$numbers = [];  for (\$i = \$min; \$i <= \$max; \$i++) {  for (\$j = \$min; \$j <= \$max; \$j++) {     \$numbers[] = pow(\$i, \$j);  } }  \$unique_numbers = array_unique(\$numbers); sort(\$unique_numbers);  return \$unique_numbers;}echo print_r(distinctPowers(2, 5), 1); // [4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125]echo print_r(count(distinctPowers(2, 100)), 1); // 9183 distinct terms`

The Kaprekars Constant problem is a little bit harder to solve. It’s the first problem in this list that requires recursion to solve the problem.

`function KaprekarsConstant(\$number, \$numberOfIterations = 1) {  \$number = (string) \$number;    if (strlen(\$number) < 4) {    for (\$i = strlen(\$number); \$i < 4; \$i++) {      \$number .= '0';    }  }    \$asc = str_split(\$number);  \$desc = \$asc;    rsort(\$desc);  sort(\$asc);    \$asc_number = (int) implode(\$asc, '');  \$desc_number = (int) implode(\$desc, '');  \$difference = abs(\$asc_number - \$desc_number);    if (\$difference !== 6174) {    return KaprekarsConstant(\$difference, \$numberOfIterations + 1);  }    return \$numberOfIterations; }echo KaprekarsConstant(2111); // 5echo KaprekarsConstant(9831); // 7`

This one took me a while to figure out. The trick in my solution is to pass variables by reference instead of by value. Still, this one might take some time to get your head around.

`function swapPairs(\$head) {    \$current = &\$head;        while (!is_null(\$current->next)) {        \$nextValue = \$current->next->val;                \$temp = &\$current;        \$temp->next->val = \$temp->val;        \$temp->val = \$nextValue;                \$current = &\$current->next->next;    }        return \$head;}`

# Want to Solve More Programming Problems Yourself?

If you enjoyed solving these problems, go to one of the websites that I mentioned in this piece. Most websites provide plenty of free challenges that you can try to solve.

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## Daan

#### Backend developer from The Netherlands. Crypto enthusiast. 