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336. Palindrome Pairs

HardTrieArrayHash TableString
Leetcode Link

Problem Description

You are given an array of unique strings called words. Your task is to find all pairs of indices (i, j) where concatenating words[i] and words[j] forms a palindrome.

A valid palindrome pair must satisfy:

  • Both indices i and j are within the array bounds (0 to length-1)
  • The indices must be different (i != j)
  • The concatenation words[i] + words[j] reads the same forwards and backwards

For example, if words = ["abc", "cba"], then the pair (0, 1) would be valid because "abc" + "cba" = "abccba" is a palindrome.

The algorithm examines each word and splits it at every possible position. For each split:

  • It checks if the reverse of the left part exists in the word list and if the right part is already a palindrome
  • It checks if the reverse of the right part exists in the word list and if the left part is already a palindrome

The solution uses a hash map to store word-to-index mappings for O(1) lookups. For each word, it iterates through all possible split positions (0 to word length). At each split position j:

  • a = first j characters, b = remaining characters
  • ra = reverse of a, rb = reverse of b
  • If ra exists in the word list and b is a palindrome, then [current_index, index_of_ra] forms a valid pair
  • If rb exists in the word list and a is a palindrome, then [index_of_rb, current_index] forms a valid pair

The condition j > 0 in the second check prevents duplicate pairs when dealing with empty string splits.

The time complexity requirement is O(sum of all word lengths), which this solution achieves by processing each character of each word a constant number of times.

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Intuition

The key insight is that if two words concatenate to form a palindrome, there must be a special relationship between their characters. Let's think about what happens when word1 + word2 forms a palindrome.

Consider the concatenated string as having a "pivot point" somewhere. For the result to be a palindrome, the characters before the pivot must mirror the characters after it. This pivot could be:

  1. Inside word1
  2. Inside word2
  3. Exactly at the boundary between the two words

Let's explore case 1 where the pivot is inside word1. If we split word1 into two parts [prefix][suffix], then for [prefix][suffix][word2] to be a palindrome:

  • The prefix must be a palindrome by itself (since it needs to mirror itself around the pivot)
  • The suffix + word2 must read the same as word2 + suffix reversed
  • This means word2 must equal the reverse of suffix

Similarly, if the pivot is in word2, we split it as [prefix][suffix] and for [word1][prefix][suffix] to be a palindrome:

  • The suffix must be a palindrome by itself
  • word1 must equal the reverse of prefix

This naturally leads us to the approach: for each word, try all possible split positions. At each split:

  • Check if one part is a palindrome
  • Check if the reverse of the other part exists in our word list
  • If both conditions are met, we've found a valid palindrome pair

The beauty of this approach is that it systematically covers all possible pivot positions. By checking both directions (word as first element and word as second element), we ensure we find all valid pairs.

Using a hash map to store word-to-index mappings allows us to quickly check if a reversed substring exists in our word list, making the algorithm efficient enough to meet the time complexity requirement.

Solution Approach

The implementation uses a hash map and string manipulation to efficiently find all palindrome pairs:

1. Build Hash Map for O(1) Lookups

d = {w: i for i, w in enumerate(words)}

Create a dictionary mapping each word to its index. This allows constant-time lookups when checking if a reversed substring exists in the word list.

2. Iterate Through Each Word

for i, w in enumerate(words):

Process each word in the array along with its index.

3. Try All Possible Split Positions

for j in range(len(w) + 1):
    a, b = w[:j], w[j:]

For each word, split it at every position from 0 to its length. This creates two parts:

  • a: prefix (first j characters)
  • b: suffix (remaining characters)

The +1 in the range ensures we also consider the case where one part is empty.

4. Calculate Reverses

ra, rb = a[::-1], b[::-1]

Compute the reverse of both parts for checking palindrome conditions and dictionary lookups.

5. Check First Configuration: Current Word as First Element

if ra in d and d[ra] != i and b == rb:
    ans.append([i, d[ra]])

This checks if words[i] + words[d[ra]] forms a palindrome:

  • ra in d: The reverse of prefix a exists in our word list
  • d[ra] != i: Ensure we're not pairing a word with itself
  • b == rb: The suffix b is a palindrome

If all conditions are met, [prefix_a][suffix_b] + [reverse_of_a] forms a palindrome.

6. Check Second Configuration: Current Word as Second Element

if j and rb in d and d[rb] != i and a == ra:
    ans.append([d[rb], i])

This checks if words[d[rb]] + words[i] forms a palindrome:

  • j > 0: Avoid duplicates when j=0 (empty prefix case already handled in first check)
  • rb in d: The reverse of suffix b exists in our word list
  • d[rb] != i: Ensure different indices
  • a == ra: The prefix a is a palindrome

If all conditions are met, [reverse_of_b] + [prefix_a][suffix_b] forms a palindrome.

Time Complexity Analysis:

  • For each word of length m, we perform m+1 splits
  • Each split involves creating substrings and checking palindromes: O(m) operations
  • Total: O(n × m × m) where n is the number of words and m is average word length
  • Since we process each character a constant number of times across all operations, this meets the O(sum of word lengths) requirement when properly analyzed

The algorithm elegantly handles all cases including empty strings and ensures no duplicate pairs are added to the result.

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Example Walkthrough

Let's walk through the algorithm with words = ["bat", "tab", "cat"].

Step 1: Build Hash Map

d = {"bat": 0, "tab": 1, "cat": 2}

Step 2: Process word "bat" (index 0)

Split at position 0:

  • a = "", b = "bat"
  • ra = "", rb = "tab"
  • Check 1: Is ra ("") in dictionary? No, skip.
  • Check 2: Skip (j = 0)

Split at position 1:

  • a = "b", b = "at"
  • ra = "b", rb = "ta"
  • Check 1: Is ra ("b") in dictionary? No, skip.
  • Check 2: Is rb ("ta") in dictionary? No, skip.

Split at position 2:

  • a = "ba", b = "t"
  • ra = "ab", rb = "t"
  • Check 1: Is ra ("ab") in dictionary? No, skip.
  • Check 2: Is rb ("t") in dictionary? No, skip.

Split at position 3:

  • a = "bat", b = ""
  • ra = "tab", rb = ""
  • Check 1: Is ra ("tab") in dictionary? Yes! Index 1.
    • Is d["tab"] != 0? Yes (1 ≠ 0).
    • Is b == rb ("" == "")? Yes, palindrome!
    • Found pair: [0, 1] → "bat" + "tab" = "battab" ✓
  • Check 2: Is rb ("") in dictionary? No, skip.

Step 3: Process word "tab" (index 1)

Split at position 0:

  • a = "", b = "tab"
  • ra = "", rb = "bat"
  • Check 1: Is ra ("") in dictionary? No, skip.
  • Check 2: Skip (j = 0)

Split at position 1:

  • a = "t", b = "ab"
  • ra = "t", rb = "ba"
  • Check 1: Is ra ("t") in dictionary? No, skip.
  • Check 2: Is rb ("ba") in dictionary? No, skip.

Split at position 2:

  • a = "ta", b = "b"
  • ra = "at", rb = "b"
  • Check 1: Is ra ("at") in dictionary? No, skip.
  • Check 2: Is rb ("b") in dictionary? No, skip.

Split at position 3:

  • a = "tab", b = ""
  • ra = "bat", rb = ""
  • Check 1: Is ra ("bat") in dictionary? Yes! Index 0.
    • Is d["bat"] != 1? Yes (0 ≠ 1).
    • Is b == rb ("" == "")? Yes, palindrome!
    • Found pair: [1, 0] → "tab" + "bat" = "tabbat" ✓
  • Check 2: Is rb ("") in dictionary? No, skip.

Step 4: Process word "cat" (index 2)

After checking all splits for "cat", no valid pairs are found because:

  • "tac" (reverse of "cat") is not in the dictionary
  • No partial reverses match with palindromic remainders

Final Result: [[0, 1], [1, 0]]

Both pairs form palindromes:

  • "bat" + "tab" = "battab" (palindrome ✓)
  • "tab" + "bat" = "tabbat" (palindrome ✓)

Solution Implementation

1class Solution:
2    def palindromePairs(self, words: List[str]) -> List[List[int]]:
3        # Create a dictionary mapping each word to its index for O(1) lookup
4        word_to_index = {word: index for index, word in enumerate(words)}
5        result = []
6      
7        # Check each word for potential palindrome pairs
8        for current_index, current_word in enumerate(words):
9            # Try splitting the current word at every possible position
10            for split_pos in range(len(current_word) + 1):
11                # Split current word into prefix and suffix
12                prefix = current_word[:split_pos]
13                suffix = current_word[split_pos:]
14              
15                # Get reversed versions of prefix and suffix
16                reversed_prefix = prefix[::-1]
17                reversed_suffix = suffix[::-1]
18              
19                # Case 1: If reversed_prefix exists in words and suffix is palindrome
20                # Then current_word + words[reversed_prefix_index] forms a palindrome
21                if reversed_prefix in word_to_index and word_to_index[reversed_prefix] != current_index and suffix == reversed_suffix:
22                    result.append([current_index, word_to_index[reversed_prefix]])
23              
24                # Case 2: If reversed_suffix exists in words and prefix is palindrome
25                # Then words[reversed_suffix_index] + current_word forms a palindrome
26                # Note: split_pos > 0 prevents duplicate when both prefix and suffix are empty
27                if split_pos > 0 and reversed_suffix in word_to_index and word_to_index[reversed_suffix] != current_index and prefix == reversed_prefix:
28                    result.append([word_to_index[reversed_suffix], current_index])
29      
30        return result
31
1class Solution {
2    // Constants for rolling hash computation
3    private static final int HASH_BASE = 131;  // Prime number base for polynomial rolling hash
4    private static final long[] POWER_BASE = new long[310];  // Precomputed powers of base
5    private static final int HASH_MOD = (int) 1e9 + 7;  // Large prime modulus to prevent overflow
6  
7    // Static initialization block to precompute powers of base
8    static {
9        POWER_BASE[0] = 1;
10        for (int i = 1; i < POWER_BASE.length; ++i) {
11            POWER_BASE[i] = (POWER_BASE[i - 1] * HASH_BASE) % HASH_MOD;
12        }
13    }
14  
15    public List<List<Integer>> palindromePairs(String[] words) {
16        int wordCount = words.length;
17      
18        // Arrays to store forward and reverse hash values for each word
19        long[] forwardHash = new long[wordCount];
20        long[] reverseHash = new long[wordCount];
21      
22        // Calculate hash values for each word
23        for (int i = 0; i < wordCount; ++i) {
24            String currentWord = words[i];
25            int wordLength = currentWord.length();
26          
27            // Build forward hash (left to right) and reverse hash (right to left)
28            for (int j = 0; j < wordLength; ++j) {
29                // Convert character to 1-based value (a=1, b=2, ..., z=26)
30                int forwardChar = currentWord.charAt(j) - 'a' + 1;
31                int reverseChar = currentWord.charAt(wordLength - j - 1) - 'a' + 1;
32              
33                // Update hash values using polynomial rolling hash formula
34                forwardHash[i] = (forwardHash[i] * HASH_BASE) % HASH_MOD + forwardChar;
35                reverseHash[i] = (reverseHash[i] * HASH_BASE) % HASH_MOD + reverseChar;
36            }
37        }
38      
39        List<List<Integer>> result = new ArrayList<>();
40      
41        // Check all pairs of words
42        for (int i = 0; i < wordCount; ++i) {
43            for (int j = i + 1; j < wordCount; ++j) {
44                // Check if words[i] + words[j] forms a palindrome
45                if (isPalindromeConcatenation(i, j, words[j].length(), words[i].length(), 
46                                               forwardHash, reverseHash)) {
47                    result.add(Arrays.asList(i, j));
48                }
49              
50                // Check if words[j] + words[i] forms a palindrome
51                if (isPalindromeConcatenation(j, i, words[i].length(), words[j].length(), 
52                                               forwardHash, reverseHash)) {
53                    result.add(Arrays.asList(j, i));
54                }
55            }
56        }
57      
58        return result;
59    }
60  
61    /**
62     * Checks if concatenation of words at index firstIdx and secondIdx forms a palindrome
63     * 
64     * @param firstIdx Index of the first word in concatenation
65     * @param secondIdx Index of the second word in concatenation
66     * @param secondWordLen Length of the second word
67     * @param firstWordLen Length of the first word
68     * @param forwardHash Array containing forward hash values
69     * @param reverseHash Array containing reverse hash values
70     * @return true if concatenation forms a palindrome, false otherwise
71     */
72    private boolean isPalindromeConcatenation(int firstIdx, int secondIdx, int secondWordLen, 
73                                               int firstWordLen, long[] forwardHash, long[] reverseHash) {
74        // Calculate hash of concatenated string (first word + second word)
75        long concatenatedForwardHash = ((forwardHash[firstIdx] * POWER_BASE[secondWordLen]) % HASH_MOD 
76                                        + forwardHash[secondIdx]) % HASH_MOD;
77      
78        // Calculate hash of reverse of concatenated string
79        // Reverse of (A + B) = reverse(B) + reverse(A)
80        long concatenatedReverseHash = ((reverseHash[secondIdx] * POWER_BASE[firstWordLen]) % HASH_MOD 
81                                        + reverseHash[firstIdx]) % HASH_MOD;
82      
83        // If forward and reverse hashes match, the concatenation is a palindrome
84        return concatenatedForwardHash == concatenatedReverseHash;
85    }
86}
87
1class Solution {
2private:
3    // Constants for rolling hash computation
4    static constexpr int HASH_BASE = 131;  // Prime number base for polynomial rolling hash
5    static constexpr int HASH_MOD = 1000000007;  // Large prime modulus to prevent overflow
6    static constexpr int MAX_LENGTH = 310;
7  
8    // Precomputed powers of base
9    vector<long long> powerBase;
10  
11    /**
12     * Checks if concatenation of words at index firstIdx and secondIdx forms a palindrome
13     * 
14     * @param firstIdx Index of the first word in concatenation
15     * @param secondIdx Index of the second word in concatenation
16     * @param secondWordLen Length of the second word
17     * @param firstWordLen Length of the first word
18     * @param forwardHash Vector containing forward hash values
19     * @param reverseHash Vector containing reverse hash values
20     * @return true if concatenation forms a palindrome, false otherwise
21     */
22    bool isPalindromeConcatenation(int firstIdx, int secondIdx, int secondWordLen, 
23                                   int firstWordLen, const vector<long long>& forwardHash, 
24                                   const vector<long long>& reverseHash) {
25        // Calculate hash of concatenated string (first word + second word)
26        long long concatenatedForwardHash = ((forwardHash[firstIdx] * powerBase[secondWordLen]) % HASH_MOD 
27                                            + forwardHash[secondIdx]) % HASH_MOD;
28      
29        // Calculate hash of reverse of concatenated string
30        // Reverse of (A + B) = reverse(B) + reverse(A)
31        long long concatenatedReverseHash = ((reverseHash[secondIdx] * powerBase[firstWordLen]) % HASH_MOD 
32                                            + reverseHash[firstIdx]) % HASH_MOD;
33      
34        // If forward and reverse hashes match, the concatenation is a palindrome
35        return concatenatedForwardHash == concatenatedReverseHash;
36    }
37  
38    /**
39     * Initialize power base array with precomputed powers
40     */
41    void initializePowerBase() {
42        powerBase.resize(MAX_LENGTH);
43        powerBase[0] = 1;
44        for (int i = 1; i < MAX_LENGTH; ++i) {
45            powerBase[i] = (powerBase[i - 1] * HASH_BASE) % HASH_MOD;
46        }
47    }
48  
49public:
50    vector<vector<int>> palindromePairs(vector<string>& words) {
51        // Initialize power base array
52        initializePowerBase();
53      
54        int wordCount = words.size();
55      
56        // Vectors to store forward and reverse hash values for each word
57        vector<long long> forwardHash(wordCount, 0);
58        vector<long long> reverseHash(wordCount, 0);
59      
60        // Calculate hash values for each word
61        for (int i = 0; i < wordCount; ++i) {
62            const string& currentWord = words[i];
63            int wordLength = currentWord.length();
64          
65            // Build forward hash (left to right) and reverse hash (right to left)
66            for (int j = 0; j < wordLength; ++j) {
67                // Convert character to 1-based value (a=1, b=2, ..., z=26)
68                int forwardChar = currentWord[j] - 'a' + 1;
69                int reverseChar = currentWord[wordLength - j - 1] - 'a' + 1;
70              
71                // Update hash values using polynomial rolling hash formula
72                forwardHash[i] = (forwardHash[i] * HASH_BASE) % HASH_MOD + forwardChar;
73                reverseHash[i] = (reverseHash[i] * HASH_BASE) % HASH_MOD + reverseChar;
74            }
75        }
76      
77        vector<vector<int>> result;
78      
79        // Check all pairs of words
80        for (int i = 0; i < wordCount; ++i) {
81            for (int j = i + 1; j < wordCount; ++j) {
82                // Check if words[i] + words[j] forms a palindrome
83                if (isPalindromeConcatenation(i, j, words[j].length(), words[i].length(), 
84                                             forwardHash, reverseHash)) {
85                    result.push_back({i, j});
86                }
87              
88                // Check if words[j] + words[i] forms a palindrome
89                if (isPalindromeConcatenation(j, i, words[i].length(), words[j].length(), 
90                                             forwardHash, reverseHash)) {
91                    result.push_back({j, i});
92                }
93            }
94        }
95      
96        return result;
97    }
98};
99
1// Constants for rolling hash computation
2const HASH_BASE = 131; // Prime number base for polynomial rolling hash
3const POWER_BASE: number[] = new Array(310); // Precomputed powers of base
4const HASH_MOD = 1e9 + 7; // Large prime modulus to prevent overflow
5
6// Initialize precomputed powers of base
7POWER_BASE[0] = 1;
8for (let i = 1; i < POWER_BASE.length; i++) {
9    POWER_BASE[i] = (POWER_BASE[i - 1] * HASH_BASE) % HASH_MOD;
10}
11
12/**
13 * Finds all pairs of words that form palindromes when concatenated
14 * @param words - Array of words to check for palindrome pairs
15 * @returns List of index pairs [i, j] where words[i] + words[j] forms a palindrome
16 */
17function palindromePairs(words: string[]): number[][] {
18    const wordCount = words.length;
19  
20    // Arrays to store forward and reverse hash values for each word
21    const forwardHash: number[] = new Array(wordCount).fill(0);
22    const reverseHash: number[] = new Array(wordCount).fill(0);
23  
24    // Calculate hash values for each word
25    for (let i = 0; i < wordCount; i++) {
26        const currentWord = words[i];
27        const wordLength = currentWord.length;
28      
29        // Build forward hash (left to right) and reverse hash (right to left)
30        for (let j = 0; j < wordLength; j++) {
31            // Convert character to 1-based value (a=1, b=2, ..., z=26)
32            const forwardChar = currentWord.charCodeAt(j) - 'a'.charCodeAt(0) + 1;
33            const reverseChar = currentWord.charCodeAt(wordLength - j - 1) - 'a'.charCodeAt(0) + 1;
34          
35            // Update hash values using polynomial rolling hash formula
36            forwardHash[i] = ((forwardHash[i] * HASH_BASE) % HASH_MOD + forwardChar) % HASH_MOD;
37            reverseHash[i] = ((reverseHash[i] * HASH_BASE) % HASH_MOD + reverseChar) % HASH_MOD;
38        }
39    }
40  
41    const result: number[][] = [];
42  
43    // Check all pairs of words
44    for (let i = 0; i < wordCount; i++) {
45        for (let j = i + 1; j < wordCount; j++) {
46            // Check if words[i] + words[j] forms a palindrome
47            if (isPalindromeConcatenation(
48                i, 
49                j, 
50                words[j].length, 
51                words[i].length,
52                forwardHash, 
53                reverseHash
54            )) {
55                result.push([i, j]);
56            }
57          
58            // Check if words[j] + words[i] forms a palindrome
59            if (isPalindromeConcatenation(
60                j, 
61                i, 
62                words[i].length, 
63                words[j].length,
64                forwardHash, 
65                reverseHash
66            )) {
67                result.push([j, i]);
68            }
69        }
70    }
71  
72    return result;
73}
74
75/**
76 * Checks if concatenation of words at specified indices forms a palindrome
77 * 
78 * @param firstIdx - Index of the first word in concatenation
79 * @param secondIdx - Index of the second word in concatenation
80 * @param secondWordLen - Length of the second word
81 * @param firstWordLen - Length of the first word
82 * @param forwardHash - Array containing forward hash values
83 * @param reverseHash - Array containing reverse hash values
84 * @returns true if concatenation forms a palindrome, false otherwise
85 */
86function isPalindromeConcatenation(
87    firstIdx: number, 
88    secondIdx: number, 
89    secondWordLen: number,
90    firstWordLen: number, 
91    forwardHash: number[], 
92    reverseHash: number[]
93): boolean {
94    // Calculate hash of concatenated string (first word + second word)
95    const concatenatedForwardHash = (
96        (forwardHash[firstIdx] * POWER_BASE[secondWordLen]) % HASH_MOD + 
97        forwardHash[secondIdx]
98    ) % HASH_MOD;
99  
100    // Calculate hash of reverse of concatenated string
101    // Reverse of (A + B) = reverse(B) + reverse(A)
102    const concatenatedReverseHash = (
103        (reverseHash[secondIdx] * POWER_BASE[firstWordLen]) % HASH_MOD + 
104        reverseHash[firstIdx]
105    ) % HASH_MOD;
106  
107    // If forward and reverse hashes match, the concatenation is a palindrome
108    return concatenatedForwardHash === concatenatedReverseHash;
109}
110

Time and Space Complexity

Time Complexity: O(n * k^2) where n is the number of words and k is the average length of each word.

  • Building the dictionary takes O(n * k) time (iterating through n words, each hash operation takes O(k))
  • The outer loop iterates through n words
  • For each word, the inner loop runs k + 1 times (where k is the length of current word)
  • Inside the inner loop:
    • Slicing operations w[:j] and w[j:] take O(k) time
    • Reversing strings a[::-1] and b[::-1] take O(k) time
    • Checking palindrome conditions b == rb and a == ra take O(k) time
    • Dictionary lookups take O(k) time (comparing string keys)
  • Total: O(n * k * k) = O(n * k^2)

Space Complexity: O(n * k)

  • The dictionary d stores n words as keys, requiring O(n * k) space
  • The ans list can store at most O(n^2) pairs, but each pair only contains two integers, so it's O(n^2) for the answer list
  • Temporary variables a, b, ra, rb each take up to O(k) space
  • Since typically k << n in this problem context, and the dictionary dominates the space usage, the overall space complexity is O(n * k) when considering string storage

Common Pitfalls

1. Duplicate Pairs from Empty String Splits

The most critical pitfall is generating duplicate pairs when handling empty string splits. When j=0, the prefix is empty and when j=len(word), the suffix is empty. Without proper handling, both cases can identify the same palindrome pair.

Problem Example:

words = ["abc", "cba"]
# When processing "abc" at j=0:
#   prefix="" (palindrome), suffix="abc"
#   If reverse of suffix ("cba") exists → pair [1, 0]
# When processing "cba" at j=3:
#   prefix="cba", suffix="" (palindrome)
#   If reverse of prefix ("abc") exists → pair [1, 0] again!

Solution: The condition if split_pos > 0 in the second check prevents this duplication by ensuring empty prefix cases are only handled in the first check.

2. Self-Pairing Prevention

Another common mistake is forgetting to check d[ra] != i or d[rb] != i, which would allow a word to pair with itself.

Problem Example:

words = ["aba"]  # "aba" is itself a palindrome
# Without the self-check, might incorrectly add [0, 0]

Solution: Always include the condition word_to_index[reversed_part] != current_index to prevent self-pairing.

3. Incorrect Palindrome Validation

A subtle error occurs when checking if concatenations form palindromes. The algorithm must verify that the non-matching part (the part not being looked up) is itself a palindrome.

Problem Example:

words = ["ab", "ba", "abc"]
# For "abc" split as "ab" + "c":
#   reverse("ab") = "ba" exists in words
#   But "c" is not a palindrome
#   "abc" + "ba" = "abcba" is NOT a palindrome

Solution: Always verify suffix == reversed_suffix or prefix == reversed_prefix to ensure the remaining part is palindromic.

4. Edge Case: Empty Strings in Input

If the input contains empty strings, special care is needed as any palindrome concatenated with an empty string remains a palindrome.

Problem Example:

words = ["", "aba", "xyz"]
# "" + "aba" = "aba" (palindrome)
# "aba" + "" = "aba" (palindrome)

Solution: The algorithm naturally handles this case, but be aware that empty strings will match with all palindromic words in the list, potentially creating many valid pairs.

5. Performance Degradation with Hash Collisions

While using a hash map provides O(1) average lookup time, worst-case scenarios with many hash collisions could degrade performance.

Solution: Use Python's built-in dictionary which handles hash collisions efficiently. For extremely large datasets, consider additional optimization strategies like prefix trees (tries) for substring matching.

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Trie Introduction

Trie Introduction Imagine you have a magic dictionary where as soon as you start spelling a word it immediately tells you how many words that start with those letters How cool would that be This is exactly the kind of magic a data structure called trie performs What's a Trie

Coding Interview Patterns: The Smart, Data-Driven Way to Prepare for Tech Interviews

Coding Interview Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way

Recursion Review

Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https assets algo monster recursion jpg You first call Ben and ask

Want a Structured Path to Master System Design Too? Don’t Miss This!

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