Hints for ~250 leetcode problems. Mostly medium, some hard. Good practice list for internships/job hunt.
If this helps you, please star the repo (I commit a lot, so I suggest you unwatch the repo).
Solutions here
- BF - Brute Force, many times I use Array and BF interchangeably; just implement the obvious logic.
- DP - Dynamic Programming
- Array - Problem specific array logic
- D&C - Divide & Conquer
| # | Tags | Hints | Complexity |
|---|---|---|---|
| 1 | Hash | Use hash map to store all values until current value. | O(n) |
| 2 | BF | Brute force. | O(n) |
| 3 | Array | Use a window. Extend if next element is not in the current window. | O(n) |
| 5 | DP | Build recurrence relation (may help to solve longest common substring problem) | O(n^2) |
| 7 | BF | Brute force. | O(log(n)) |
| 9 | BF | Brute force. | O(log(n)) |
| 10 | DP | Use a 2D matrix. DP[i][j] = whether p[0:i] matches with str[0:j]. | O(mn) |
| 15 | Array | Solve 2 sum before attempting this problem. | O(n^2) |
| 16 | Array | Similar to 3 sum, but use a sorted array and make front++, back-- updates to minimize abs(target - sum). | O(n^2) |
| 17 | Recursion | Possible approaches are trie, or more easy - building on smaller solution. | O(4^n) |
| 22 | Recursion | If less than n open brackets are open, we can add open bracket. But to add a close bracket you need count(')') < count('(') | . |
| 23 | Heap | Use a data structure which can give you minimum element at low cost. | . |
| 26 | Array | Maintain length of unique elements till current element. | O(n) |
| 27 | Array | vector.erase() will not work, as vector overwrites the element every time erase is called. | O(n) |
| 28 | BF | Brute force. | O(mn) |
| 31 | Array, Sorting | Find the location closest from the end which has segment ab of the form a < b, and make it ba. | O(n log(n)) |
| 33 | Binary Search | Array increasing, sudden drop and then increasing again. Find out where the middle element is, can it's location help you? | O(log(n)) |
| 36 | BF | Brute force. | O(1) |
| 42 | Stack + Array | Find next greater element index for each index, and prefix sum for each element. Can you build the solution from this? Check out problem 496 for next greater element. | O(n) |
| 43 | BF | Normal multiplication, like in school. | O(n log(d)) |
| 44 | DP | Solve problem 10 first. Construct a 2D box of size len(pattern)*len(string). DP[i][j] = whether p[0:i] matches with str[0:j] | O(mn) |
| 46 | Recursion | Combine str[0] with permutations of str[1:], use recursion. | O(n!) |
| 47 | Recursion | Eg, aaaabbcc, only one of each kind {a, b, c} can come at 0th index. Recurse. | O(n!) |
| 48 | Array | index <i, j> is i positions from left, j positions from top. Find a mapping to 3 other locations, and swap the values. | O(n^2) |
| 49 | Hash | Hashing | O(n) if O(1) hash |
| 50 | D&C/Recursion | 2^n = 2^(n/2)*2^(n/2) | O(log(n)) |
| 51 | Backtracking + Pruning | If a queen is in <i, j>, block that diagonal, row & column. | O(n!) |
| 52 | Backtracking | Solution for 51 is more exhaustive than this problem. | O(n!) |
| 53 | Array | Keep extending window <i, j> to the right until sum > 0, if sum < 0, set i = j+1 | O(n) |
| 54 | Array/BF | Brute Force | O(n^2) |
| 56 | Sorting | Sort based on start time. | O(n log(n)) |
| 57 | Sorting | Insert interval and do problem 56. | O(n log(n)) |
| 66 | BF | Brute force | O(n) |
| 67 | BF | Simple array addition | O(log(n)) |
| 68 | Array | If you can fit in k words in a line, there must by atleast k-1 spaces, so L-(k-1)-sum(char's in words) must be divided mentioned in the question. | O(n) |
| 69 | Binary Search | start = 0, end = x/2, binary search | O(log(n)) |
| 70 | DP | DP[i] = DP[i-1] + DP[i-2] | O(n) |
| 71 | BF | Split by '/', use stack to handle '..' | O(n) |
| 72 | DP | 2D DP, DP[i][j] = minimum steps for string1 till i to match with string2 until j. Also note that adding a character in one string is the same as removing it in another. | O(n^2) |
| 75 | 2 pointer | Start & end indices. See if you can swap em into place. | O(n) |
| 76 | Array | Extend and contract window. | O(n) |
| 77 | Backtrack | Backtracking | O(nCk) |
| 78 | Backtrack | Backtracking Or generate all numbers between 0 & 2^n and use bit properties. | O(2^n) |
| 79 | DFS | DFS if letter match | O(nm * len(str)) |
| 80 | BF | Implementation based, maintain count. | O(n) |
| 81 | Binary Search | Similar to Problem 33. | O(log(n)) |
| 88 | Array | Merging logic from merge sort | O(n+m) |
| 90 | Array | Build the powerset starting from []. [[], [1]] + [2] -> [[], [1], [2], [1, 2]] + [2] = [[], [1], [2], [1, 2], [2, 2], [1, 2, 2]} | O(2^n) |
| 91 | DP | DP[i] = number of ways to decode until position i, relate it with DP[i-1] & DP[i-1] | O(n) |
| 98 | Tree Traversal | Both subtrees must be BST and max(left subtree) < min(right subtree) | O(n) |
| 101 | Tree Traversal | Assume you're at root, ll - left subtree of left subtree, and so on; Condition is ll = rr & lr = rl & l->val = r->val. | O(n) |
| 102 | BFS | Straighforward BFS | O(n) |
| 103 | BFS | BFS + stack at alternate levels. | O(n) |
| 104 | Tree Traversal | f(i) = 1 + max(f(i->left), f(i->right)) | O(n) |
| 105 | |||
| 106 | |||
| 107 | Tree Traversal | Problem 102 + reverse the result. | O(n) |
| 108 | Array | Middle element becoems root. | O(n) |
| 116 | BFS | BFS will give the ordering. | O(n) |
| 117 | BFS | Same hint as 116. | O(n) |
| 118 | BF | Generate new row from old row. | O(n^2) |
| 119 | BF | Generate new row from old row. | O(n^2) |
| 128 | DP | DP problem, find recurrence relation. | O(n) |
| 133 | DFS/BF | Copy the original graph in a DFS like manner. | O(E+V) |
| 137 | Bit manipulation/Array | Consider 1 bit version, only 0's and 1's are allowed. If all numbers except one come 3 times, then there will be either 3k 1's, or 3k 0's followed by the bit of the actual number. So, whenever 1 comes 3 times make it 0. In the end, only the bit in the required number will be existing. Extend this logic to 32 bits. | O(n) |
| 139 | DP | DP[i] = whether sentence S[1:i] has a valid word break. DP[i] is true if some DP[j] is true and S[j:i] is a word. | O(n^2) |
| 140 | DP | Same as 139, keep track of all possible solutions at each i. | O(n^2) |
| 146 | Hashing/Array | Mostly brute force; keep track of which element needs to get evicted next, elements are stored in a hash. | O(1) per operation |
| 149 | Array | For each point X, check slope with every other point. If 5 points have same slope S then they all fall on the same line where X defines the intercept for slope S. | O(n^2) |
| 150 | Stack | Use stack to store operands and whenever operator is seen, evaluate. Use stoi() to convert string to number. | O(n) |
| 152 | Array | Maintain max and min values while iterating through the array. If A[i] is -ve, then swap max and min, as min is now a candidate to become max (after multiplying). While multiplying with A[i], only multiply if it can increase max or decrease min. If it cannot, then that subarray ends there and must start with A[i]. | O(n) |
| 161 | BF | When there is mismatch, check if A[i+1:] == B[j:], A[i+1:] == B[j+1:], A[i:] == B[j+1:]. | O(n) |
| 164 | Bucketing | Try to bucket numbers b/w min and max with gap = ceil(max-min / n-1). This way minimum gap is atleast according to the formula. If the numbers are in the same bucket then they cannot form the answer. So for every bucket you only have to maintain the min and max, compare max[i-1] with min[i] for all i, as they would be consecutive in the final sorting order. Note that maybe you can't place largest number inside some bucket, so keep an extra check later. | O(n) |
| 168 | BF | If n <= 26, then this is easy. Otherwise it is F(n) = F(n/26)+ (char)n%26; something like this. Calculate the exact relation. | O(log(n)) |
| 169 | Array | Assume A[0] is the majority element i.e. curmax = 0, whenever A[i] != A[curmax], reduce the count, if the count becomes -ve at any point, set curmax = i. Idea is that we are cancelling all non-equal elements, if some element occurs more than half times, then in the end curmax should point to that element. | O(n) |
| 172 | Math | [n/5] + [n/25] + [n/125] and so on .. where [] is floor operator. | O(log(n)) |
| 173 | Stack | For every element, go right and then left repeatedly to get the next smallest element, can storing it in a stack help? | O(n) |
| 187 | Hash | Store all 10 letter substrings in a hashtable. | O(n) |
| 190 | BF | Remember to store answer in unsigned long. Construct answer from bits. | O(log(n)) |
| 191 | BF | Straighforward. | O(log(n)) |
| 198 | DP | DP[i] = max(DP[i-2]+nums[i], DP[i-1]). | O(n) |
| 200 | DFS | Do DFS whenever you encounter a 1; #islands = # of times the DFS routine is called from main(). | O(n*m) |
| 202 | Set | Use a set to keep track of all visited numbers. | Don't know. |
| 205 | Hash | Make sure you have a 1-1 mapping. | O(n) |
| 206 | BF | Use variables prevNode, curNode, nextNode. You need to do it this step by step, decide in how to and in what order to set the said variables. | O(n) |
| 207 | DFS | You're looking for a loop. So set the nodes whom you're visiting in the current loop as gray, and if you end up visiting a gray node then there is a loop. | O(V+E) |
| 208 | Tree | Multiway tree. | O(n) for both search and insert, n = # letters |
| 209 | Array | Expanding and shrinking window approach. | O(n) |
| 210 | Topological Sorting | Make sure to check for cycles. | O(V+E) |
| 212 | Trie+DFS | Problem is that you never know when a word ends. If no word starts with prefix "ab" and your path in DFS is currently "ab" you can break the current DFS chain. See if a trie can help with this kind of pruning. | O(n^2*m^2) in the absolute worst case, all words use all the letters in the matrix. |
| 214 | BF/KMP | KMP is faster. But BF works by asking .. what is the largest prefix of the string that is also a palindrome, and thus how many characters are needed to make it into a palindrome. | O(n) |
| 215 | QuickSort | Don't complete the sort. Partition, and if there are more than k characters in the first half, search for kth largest in the first half, else search for it in the second half (subtract number of elements in the first half). | O(log(n)) on average |
| 217 | HashSet | Check if an element is already in the set, if not add it. | O(n) |
| 219 | HashMap | Similar to 217. | O(n) |
| 220 | HashMap | Divide all numbers by t. Then use a hashset to check for nums[i]-1, nums[i] and nums[i+1] | O(n) |
| 221 | DP | DP[i][j] = length of biggest square ending at i, j. Use DP[i-1][j], DP[i-1][j-1], DP[i][j-1] to get and update expression for DP[i][j]. | O(mn) |
| 224 | Stack | You need to use 2 stacks. Read internet article on expression evaluation. | O(n) |
| 227 | Stack | Similar to 224, but you need to use preference order. | O(n) |
| 228 | BF | Single pass through the numbers, you should be able to get the first and the last elements of the ranges. Check what happens if there is only one element. | O(n) |
| 230 | Inorder Traversal | Find the kth largest element during traversal. | O(n) |
| 231 | Bit manipulation | Only one bit must be set. | O(1) |
| 234 | List Reversal | Reverse first half of linked list and then go outwards from the centre. | O(n) |
| 235 | BF | Current node should be larger than one element but smaller than one element. Find this node. | O(h) |
| 236 | BF | Find the path to each node, then compare the paths. | O(n) |
| 238 | Array | Compute prefix product array and suffix product array. | O(n) |
| 239 | Array/Stack | Solve Next greater element problem. Until nge index is outside the window, current maximum will repeat. | O(n) |
| 240 | Array | Start from the top right corner, if larger than target, go left, else go down. | O(m+n) |
| 246 | Array | Find out the mapping, 1-1, 6-9 and so on. Algo is similar to checking palindrome, instead check for this mapping. | O(n) |
| 247 | Recursion | Can expand $ as it as "1$1", "0$0", "6$9", "8$8" and so on. | O(5^n) |
| 248 | Recursion | Similar to 247, just check if it is in range and add it to the list. | O(5^n) |
| 249 | Hashing | For each string S, convert it into a Key K by using a concept of offsets. | O(n), string processing is assumed to be O(1) |
| 251 | BF | Keep track of which vector and position you are at. | O(n), n = # total number of elements |
| 252 | Sorting | Sort by start or end time, and check if a previous interval interrupts the next one. | O(n log(n)) |
| 253 | Greedy/Heap | Basically interval scheduling problem. | O(n log(n)) |
| 257 | Inorder traversal | Maintain a path string and add to solution when you hit null. | O(n) |
| 259 | Sorting/2Sum | Sort and do 2 sum like algo. They asked for index triplets, so duplicates aren't an issue. | O(n^2) |
| 261 | DFS | If any chain of exploration hits an already visited node, return false. | O(n) |
| 263 | BF | Divide by 2, 3, 5. You should have 1 in the end. | O(log(n)) in terms of bits. |
| 264 | BF | There are 3 sequences 2x1, 2x2, 2x3....; 3x1, 3x2, 3x3, ...; 5x1, 5x2, 5x3 ...; Keep 3 pointers, at each step, each of these pointers will give the next candidate. The minimum candidate is the next ugly number. | O(n) |
| 265 | BF/Greedy | M[i][j] = cost of painting first i house and i is painted in color j. Fill it it in a bottom up fashion. | O(n^2) |
| 266 | BF | Keep count of letters. | O(n) |
| 269 | Topological Sort | Construct the mapping using the dictionary, then do topological sorting. | O(n*max(word_length)) |
| 270 | BST | Root to leaf search. | O(log(n)) |
| 271 | Array | Use string length to keep track. | O(n), n=total number of characters |
| 272 | Tree traversal | Given a sorted order, can you extract the k numbers in O(n)? Inorder traversal gives sorted order in O(n). To extract the k numbers, find 2 numbers - the smallest number larger than target (right index), and the largest number smaller than target (left index). Extend left and right indices until you have k numbers. | O(n) |
| 273 | BF | Write a routine that deals with the lowest 3 positions of the number, you can use this to create a billion b million c thousand d. | O(log(n)) |
| 274 | Sorting | Check for conditions from highest citation to lowest citation in sorter order. | O(n) |
| 275 | Array | Same algo as above. | O(n) |
| 277 | Array | Consider P1, P2. If P1 knows P2, P1 can't be the answer and vice versa. If both know each other or both don't know each other, both can't be the answer. Extend this idea. | O(n) |
| 278 | Binary Search | BF | o(log(n)) |
| 279 | DP | F(n) = 1 + min{i = 1..sqrt(N)} (F(n-i*i)) | O(n sqrt(n)) |
| 316 | Stack | First count number of instances of each character. Go through left to right. If the stack is empty, insert. If the character is already in the stack, continue. While the character in the stack.top is larger than current character and has more instances later, remove it. Finally add current char to the stack. Core idea is that you are trying to push the smaller characters to the left as much as possible. You only force yourself to add a larger character if it never appears again. | O(n) |
| 413 | Array | [1, 2, 3, 4, 5, 7, 9, 11]. can be written as [5, 3] i.e. 5 sequence of difference 1 and 3 sequence of difference 2, you need to figure out how many parts you can split 5 and 3 into. | O(n) |
| 694 | DFS | Keep track of the directions in which DFS proceeds in some form, maybe a string like ddr for down down right. | O(rows*cols) |
| 738 | Array | Find the first time Xi > Xi+1, Xi -= 1 and turn all Xi+k = 9, For eg, 321 becomes 299. Figure out cases like 33332. | O(n) |
| # | Hints | Complexity |
|---|---|---|
| 5 | Use DP, P<i, j> = P<i+1, j-1> && S[i] == S[j] is the recurrence relation. | O(n^2) |
| 10 | Manually solve [aab, cab], [ab, .b], [aaa, ab*a] for corner cases. | O(mn) |
| 15 | 2sum for a sorted list gives a more optimal sub routine here. | O(n^2) |
| 17 | Build solution for first k numbers, and modify solution vector for k+1 number. | O(4^n) |
| 22 | Backtracking based solution. | |
| 23 | Use heap (this method uses extra memory). | |
| 27 | Count the number of occurences in a variable - acting as offset. | |
| 31 | Sort from b onwards. Solution is done. | |
| 33 | Depending upon the location of the middle element and whether target is lower or greater than middle element, write 4 cases. | |
| 42 | Let nge(i) = k and p be prefix sum array, we can roughly say water trapped = Sigma[ (h_i x (k-i)) - (p[k]-p[i])] | |
| 44 | Write cases for when p[i] = '' and not ''. | O(mn) |
| 108 | Left node must be filled before right node. | |
| 128 | If L[k] represents the largest consecutive sequence, L[k] -> L[k-L[k]]+ L[k] |
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