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Remove All Adjacent Duplicates In String

Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them.

We repeatedly make duplicate removals on S until we no longer can.

Return the final string after all such duplicate removals have been made. It is guaranteed the answer is unique.

Example 1:

Input: "abbaca"
Output: "ca"
Explanation: 
For example, in "abbaca" we could remove "bb" since the letters are adjacent 
and equal, and this is the only possible move.  The result of this move is 
that the string is "aaca", of which only "aa" is possible, so the final 
string is "ca".

Solution

/**
 * @param {string} S
 * @return {string}
 * @time complexity: O(n) where n is the length of S
 * @space complexity: O(n)
 */
var removeDuplicates = function(S) {
    const stack = [];

    for (let c of S) {
        if (c !== stack[stack.length - 1]) {
            stack.push(c);
        } else {
            stack.pop();
        }
    }

    return stack.join('');
};

Alternative Solution

/**
 * @param {string} S
 * @return {string}
 * @time complexity: O(n) where n is the length of S
 * @space complexity: O(n)
 */
var removeDuplicatesAlternative = function(S) {
  const arr = S.split('');

  for (let i = 1; i < arr.length; i++) {
    if (arr[i] === arr[i-1]) {
      arr.splice(i-1, 2);
      i -= 2;
    }
  }

  return arr.join('');
};

Test Case

const assert = require('chai').assert;

describe('Remove All Adjacent Duplicates In String', () => {

  it('should return \'ca\' when \'abbaca\' is given', () => {
    assert.strictEqual(removeDuplicates('abbaca'), 'ca');
  });

  it('should return \'\' when \'aabbacca\' is given', () => {
    assert.strictEqual(removeDuplicates('aabbacca'), '');
  });

  it('should return \'\' when \'\' is given', () => {
    assert.strictEqual(removeDuplicates(''), '');
  });

});
  Remove All Adjacent Duplicates In String
    ✓ should return 'ca' when 'abbaca' is given
    ✓ should return '' when 'aabbacca' is given
    ✓ should return '' when '' is given


  3 passing (12ms)


Mike Mai
Mike Mai   Brooklyn, New York
I am full-stack web developer, passionate about building world class web applications. Knowledge in designing, coding, testing, and debugging. I love to solve problems.