String Interview Questions 2026

Why Strings Are So Common

String and array problems together account for roughly 35% of all coding interview questions in LeakCode's database. Strings are common because they are natural: almost every software system processes text, and string problems test pointer manipulation, data structure choice, and edge case awareness in a self-contained package.

String problems also span a wide difficulty range. Two Sum on strings (two-character frequency counting) is easy. Minimum Window Substring with multiple character constraints is hard. The same core patterns appear at every difficulty level.

Sliding Window: The Core String Pattern

The sliding window pattern solves problems asking for an optimal substring or subarray satisfying some condition: longest substring without repeating characters, minimum window containing all characters, maximum sum subarray of length k.

The template: maintain a left and right pointer. Expand right to include new characters. Contract left when the window violates the condition. Update the answer at each valid window state. This template solves the majority of sliding window problems with minor variations.

Variable vs fixed window: for fixed-size windows (k characters), advance both pointers in lockstep. For variable windows (find minimum/maximum length satisfying X), use the expand-then-contract approach.

Hashmap Frequency Patterns

A large class of string problems reduces to counting character frequencies and comparing them. Anagram detection, group anagrams, find all anagrams in a string, and valid permutation all follow this pattern. The key insight: two strings are anagrams if and only if their character frequency maps are equal.

For fixed-alphabet problems (only lowercase letters), an array of size 26 is faster than a hashmap. For problems with arbitrary characters or Unicode, use a defaultdict. Know when to use each and state the reason during your interview.

Top String Questions by Category

Sliding Window for String Problems

Sliding window is the workhorse for substring problems. Maintain a window between left and right pointers. Expand right by adding the new character; shrink left when the constraint is violated. Track window state (frequency map, distinct count, sum) incrementally rather than recomputing.

Longest substring without repeating characters: track a set of characters in the window. Expand right; when a duplicate is found, shrink left until the duplicate is removed. Maintain the max window length seen. O(n) time, O(min(charset, n)) space. The discriminating bug at L4 interviews: forgetting to remove the duplicate character from the set when shrinking left.

Palindrome Patterns

Palindromic substring problems have two main techniques: expand around center (O(n^2) time, O(1) space) and Manacher's algorithm (O(n) but rarely required). Expand around center: for each index, expand around it (odd-length) and around the gap to its right (even-length). Track the longest palindrome found.

DP approach to palindrome: dp[i][j] = true if substring i..j is a palindrome. Recurrence: dp[i][j] = (s[i] == s[j]) AND (j - i < 2 OR dp[i+1][j-1]). Both approaches are O(n^2) time; DP uses O(n^2) space, expand-around-center uses O(1). Strong candidates pick the right approach for the specific question constraints.

Parsing and Stack-Based String Problems

Many string problems reduce to stack manipulation. Valid parentheses: push opening brackets, pop and verify match on closing. Decode string ("3[a2[c]]"): parse the number, push to stack, recurse or iterate. Basic calculator with parentheses: maintain a stack of partial results.

Simplify path (Unix-like): split by /, process each component (skip empty and "."; pop on ".."; push otherwise). Reverse only the words in a string: split, reverse the array, join. Whenever you see nested structure or balanced delimiters, the stack data structure is the right reach.

Edit Distance and Sequence Alignment

Edit distance (Levenshtein) is the canonical 2D DP problem on strings. dp[i][j] = minimum edits to transform s1[0..i] into s2[0..j]. Recurrence: if s1[i] == s2[j], dp[i][j] = dp[i-1][j-1]. Else dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) for delete, insert, substitute.

Variants: longest common subsequence (similar DP, take max), one-edit distance (early termination after one difference), regex matching with star/dot (DP with multiple state transitions). These appear at Google L5+ and Meta E5+ with notable frequency in 2024-2026 reports on LeakCode.