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Is Subsequence

This question is asked by Google. Given two strings s and t, return whether or not s is a subsequence of t.

Note: You may assume both s and t only consist of lowercase characters and both have a length of at least one.

Example(s)​

Example 1:

Input: s = "abc", t = "aabbcc"
Output: true
Explanation:
Characters of "abc" appear in "aabbcc" in order:
- 'a' at position 0
- 'b' at position 2
- 'c' at position 4

Example 2:

Input: s = "cpu", t = "computer"
Output: true
Explanation:
Characters of "cpu" appear in "computer" in order:
- 'c' at position 0
- 'p' at position 1
- 'u' at position 4

Example 3:

Input: s = "xyz", t = "axbyc"
Output: false
Explanation:
Characters of "xyz" do not all appear in "axbyc":
- 'x' at position 1 ✓
- 'y' at position 3 ✓
- 'z' not found ✗

Solution​

The solution uses two pointers:

  1. Pointer for s: Tracks position in string s
  2. Pointer for t: Scans through string t
  3. Match characters: When characters match, advance pointer in s
  4. Check completion: If we match all characters in s, it's a subsequence

JavaScript Solution - Two Pointers

/**
* Check if s is a subsequence of t
* @param {string} s - Subsequence to check
* @param {string} t - String to search in
* @return {boolean} - True if s is subsequence of t
*/
function isSubsequence(s, t) {
if (s.length === 0) return true;
if (s.length > t.length) return false;

let sIndex = 0; // Pointer for string s

// Scan through string t
for (let tIndex = 0; tIndex < t.length; tIndex++) {
  // If current character in t matches current character in s
  if (t[tIndex] === s[sIndex]) {
    sIndex++;
    // If we've matched all characters in s, it's a subsequence
    if (sIndex === s.length) {
      return true;
    }
  }
}

// If we didn't match all characters, it's not a subsequence
return sIndex === s.length;
}

// Test cases
console.log('Example 1:', isSubsequence("abc", "aabbcc")); 
// true

console.log('Example 2:', isSubsequence("cpu", "computer")); 
// true

console.log('Example 3:', isSubsequence("xyz", "axbyc")); 
// false

console.log('Test 4:', isSubsequence("ace", "abcde")); 
// true
Output:
Click "Run Code" to execute the code and see the results.

Alternative Solution (Recursive)​

Here's a recursive version:

/**
 * Recursive approach
 */
function isSubsequenceRecursive(s, t) {
  // Base cases
  if (s.length === 0) return true;
  if (t.length === 0) return false;
  
  // If first characters match, check rest
  if (s[0] === t[0]) {
    return isSubsequenceRecursive(s.substring(1), t.substring(1));
  }
  
  // If don't match, skip character in t
  return isSubsequenceRecursive(s, t.substring(1));
}

Complexity​

  • Time Complexity: O(n) - Where n is the length of t. We scan through t once.
  • Space Complexity: O(1) - Only using a constant amount of extra space (two pointers).

Approach​

The solution uses two pointers:

  1. Pointer for s:

    • Tracks which character in s we're looking for
    • Only advances when we find a match

  2. Pointer for t:

    • Scans through t sequentially
    • Checks if current character matches what we need in s

  3. Match and advance:

    • When characters match, advance pointer in s
    • Continue scanning t

  4. Check completion:

    • If we match all characters in s, return true
    • If we finish scanning t without matching all, return false

Key Insights​

  • Two pointers: One for each string
  • Greedy approach: Match first occurrence of each character
  • Order matters: Characters must appear in same order
  • Not consecutive: Characters don't need to be adjacent
  • O(n) time: Single pass through t

Step-by-Step Example​

Let's trace through Example 1: s = "abc", t = "aabbcc"

s = "abc"
t = "aabbcc"
sIndex = 0

tIndex = 0: t[0] = 'a'
  t[0] === s[0]? Yes ('a' === 'a')
  sIndex = 1

tIndex = 1: t[1] = 'a'
  t[1] === s[1]? No ('a' !== 'b')
  Continue

tIndex = 2: t[2] = 'b'
  t[2] === s[1]? Yes ('b' === 'b')
  sIndex = 2

tIndex = 3: t[3] = 'b'
  t[3] === s[2]? No ('b' !== 'c')
  Continue

tIndex = 4: t[4] = 'c'
  t[4] === s[2]? Yes ('c' === 'c')
  sIndex = 3
  
  sIndex === s.length? Yes (3 === 3)
  Return true

Visual Representation​

Example 1: s = "abc", t = "aabbcc"

t:  a  a  b  b  c  c
    ↑     ↑        ↑
    Match 'a' at position 0
         Match 'b' at position 2
              Match 'c' at position 4

s:  a  b  c
    ✓  ✓  ✓
    All matched in order

Result: true

Example 3: s = "xyz", t = "axbyc"

t:  a  x  b  y  c
       ↑     ↑
       Match 'x' at position 1
            Match 'y' at position 3
                 'z' not found

s:  x  y  z
    ✓  ✓  ✗
    'z' not found

Result: false

Edge Cases​

  • s is empty: Return true (empty string is subsequence of any string)
  • s longer than t: Return false
  • s equals t: Return true
  • s is single character: Works correctly
  • t is single character: Works correctly
  • No matches: Return false

Important Notes​

  • Subsequence vs substring: Subsequence doesn't require consecutive characters
  • Order matters: Characters must appear in same order
  • Greedy matching: Match first occurrence of each character
  • O(n) time: Single pass through t
  • O(1) space: Only two pointers needed

Why Two Pointers Work​

Key insight:

  • We need to find characters of s in t in order
  • Once we match a character in s, we move to the next
  • We only scan forward in t, never backward
  • This greedy approach finds the subsequence if it exists
  • Longest Common Subsequence: Different problem (find longest)
  • Isomorphic Strings: Different problem
  • Valid Anagram: Different problem
  • Edit Distance: Different problem

Takeaways​

  • Two pointers efficiently check subsequence
  • Greedy matching finds first occurrence of each character
  • Order matters - characters must appear in same order
  • O(n) time with single pass through t
  • O(1) space using only two pointers