Code War problem

My Solution on this problem:

function numberOfPairs(gloves) {
  const glove = gloves.slice().sort();
  const pairs = [];

  for (let i = 0; i < glove.length - 1; i  ) {
    if (glove[i] == glove[i 1]) {
      pairs.push(glove[i]);
    }
  }
  return pairs.length;
}

It pass the initial test but failed the attempt/random test. 2nd test

My Second Solution, I add i . It pass the Attempt

function numberOfPairs(gloves) {
  const glove = gloves.slice().sort();
  const pairs = [];

  for (let i = 0; i < glove.length - 1; i  ) {
    if (glove[i] == glove[i 1]) {
      pairs.push(glove[i]);
      i   // How?
    }
  }
  return pairs.length;
  
}

Can you guys help me how does i fix the problem.

CodePudding user response：

Because if you found a pair, the next element is equal to the actual and you're advancing only one element in the array at a time. If you found a pair, the next element was the one found, then you have to advance another position in the array when you found it.

Say you have this array: [1, 2, 5, 8, 9, 2, 8] Let's number the lines to make it easier:

 1 function numberOfPairs(gloves) {
 2     const glove = gloves.slice().sort();
 3     const pairs = []; 
 4     for (let i = 0; i < glove.length - 1; i  ) {
 5         if (glove[i] == glove[i 1]) {
 6             pairs.push(glove[i]);
 7             i   // How?
 8         }
 9     }
10     return pairs.length;
11 }

after line 2 you will have the following sorted array:

[1, 2, 2, 5, 8, 8, 9]

with 2 pairs in it. Then you will start iteration of the array with i = 0 so in line 6 in the first iteraction glove[i] is 1 and glove[i 1] is 2. Different, so the for continues to i = 1.

Now the if in line 6 tests glove[i] with i = 1 making glove[i] = 2 and glove[i 1] also = 2 (first pair). So it enters the if and pushes the first number to pairs: 2.

Now if i isn't present in line 7, what would happen is that for would countinue to i = 2 making glove[i] = 2 and glove[i 1] = 5. But the 2 makes part of the first pair encountered. So this is incorrect. We must skip the 2nd 2 that's what i is line 7 is there for. It would be even worse if the 4th element in the array was 2 again because it would inform another pair when there only one.

After this for will continue not to i=2 but correctly to i=3 testing elements 5 and 8 for the next pair.

Hope this explains clearly enough.

CodePudding user response：

You can also group the colors and then count the pairs using Array.prototype.reduce.

function solution(input) {
  return input.reduce(
    ([pairs, hash], clr) => {
      hash[clr] = (hash[clr] ?? 0)   1;
      return [pairs   (hash[clr] % 2 === 0 ? 1 : 0), hash];
    },
    [0, {}]
  )[0];
}

console.log(solution(["red", "green", "red", "blue", "blue"]));
console.log(solution(["red", "red", "red", "red", "red", "red"]));