AKINJIDE BANKOLE


6 min read

Reducing Search Space

Madara with Nine Tails Sketch

Not long ago, I was tasked to write a program that generates offsets of any given number (i.e. n = 3) and this generated offsets (i.e. a = [2, 3, 4, 5]) will be used to intersect a much bigger list (i.e. b = [1, 2, 5, 6, 7, 8, 9]), a, intersects b returning (i.e. x = [2, 5]), X will be the result of the intersection.

Although the task took a while to accomplish, I discovered the program’s first iteration was not performant, therefore I did research on how best to make it more performant and I’d like to share the research and program iterations in this article.

First Iteration

The below code is not performant but I utilized Closures and Lexical scoping to accomplish the initial program. Even though the code below uses an Array, a similar data structure can be used in replace:

function offsetFactory(startWithNumber, count) {
  const createOffset = (number, count) => {
    let result = [];
    let head = number;
    let tail = number;

    for (let j = 0; j <= count; j += 1) {
      if (result.length + 1 === count) {
        result.push(number);
        break;
      }

      if (j % 2 === 0 && head >= 0) {
        head += 1;
        result.push(head);
      } else if (j % 2 === 1 && tail > 0) {
        tail -= 1;
        result.push(tail);
      } else {
        head += 1;
        result.push(head);
      }
    }

    return result.sort((a, b) => a - b);
  };

  const offsets = createOffset(Number(startWithNumber), parseInt(count, 10));
  console.log('offsets: ', offsets);
  // offsets: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …] (100)

  return (searchNumber) => {
    let isMatch = false;

    if (offsets && offsets.length > 0) {
      for (const offset of offsets) {
        isMatch = String(searchNumber) === String(offset);

        if (isMatch) {
          return isMatch;
        }
      }
    }

    return isMatch;
  };
};

const filterFn = offsetFactory(50, 100);
const fibSequence = [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377];

console.log('match: ', fibSequence.filter(filterFn));
// match: [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …] (11)

offsetFactory() has a function declared within, createOffset() which creates a local variable offsets and a single lexical environment that is shared by the anonymous inner function. The inner function defined inside offsetFactory() has its own local variables, however, since inner functions have access to variables of outer functions, offsets variable declared in the parent function, offsetFactory() can be accessed by the anonymous inner function.

createOffset() accepts 2 parameters, number to offset from and count indicating how many numbers to generate. The loop uses the count parameter and modulus to generate numbers on both left and right of the number provided as a parameter and afterward push the initial number provided into the result.

The anonymous inner function will be invoked as a callback function to any standard method that accepts function as an argument and the anonymous inner function accepts 1 parameter, searchNumber for filtering through the local variable offsets created by the createOffset() function. If there’s a match during an iteration of local variable offsets the anonymous inner function returns true otherwise false.

Run the code

You’ll notice fibSequence is filtered by filterFn(). Although it works as expected, the performance is not good enough. What happens here is every time filterFn() is invoked with a new searchNumber from fibSequence; closure variable, offsets is iterated and compared with searchNumber until a match is found.

Calculate the cost

If given (i.e. offsets = [1..1000]) and (i.e. fibSequence = [1..2000]), iterating through fibSequence linearly X times and for each M item, iterate over offsets linearly N items :(, sad!. Giving, M * N.

Second Iteration

After discovering this problem, my research revealed the high-cost operation can be reduced by yanking 1 item off the local variable, offsets for each fibSequence number match. The below code gives insight:

function offsetFactory(startWithNumber, count) {
  const createOffset = (number, count) => {
    let result = [];
    let head = number;
    let tail = number;

    for (let j = 0; j <= count; j += 1) {
      if (result.length + 1 === count) {
        result.push(number);
        break;
      }

      if (j % 2 === 0 && head >= 0) {
        head += 1;
        result.push(head);
      } else if (j % 2 === 1 && tail > 0) {
        tail -= 1;
        result.push(tail);
      } else {
        head += 1;
        result.push(head);
      }
    }

    return result.sort((a, b) => a - b);
  };

  const offsets = createOffset(Number(startWithNumber), parseInt(count, 10));
  console.log('offsets: ', offsets);
  // offsets: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …] (100)

  return (searchNumber) => {
    let isMatch = false;

    if (offsets && offsets.length > 0) {
      for (const offset of offsets) {
        isMatch = String(searchNumber) === String(offset);

        if (isMatch) {
          const index = offsets.indexOf(id);

          if (index > -1) {
            offsets.splice(index, 1);
          }

          return isMatch;
        }
      }
    }

    return isMatch;
  };
};

const filterFn = offsetFactory(50, 100);
const fibSequence = [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377];

console.log('match: ', fibSequence.filter(filterFn));
// match: [1, 2, 3, 5, 8, 13, 21, 34, 55, …] (11)

Run the code

You might not notice that much change but the code above is more performant compared to the previous code. What happens here is every time filterFn() is invoked with a new searchNumber from fibSequence, each local variable, offsets number is iterated and compared with searchNumber if both match we keep searchNumber and remove the matched number from the local variable, offsets Array, repeating the same process for each searchNumber.

The only problem with this optimized approach is uniqueness, fibSequence declaration above contains number 1 at index 1 and 2 and we expect final output to contain 1 twice, but this is not the case ‘cause after matching 1 at index 1, the matched number is removed from local variable, offsets and that will prevent 1 at index 2 from getting matched.

Calculate the cost

If given (i.e. offsets = [1..1000]) and (i.e. fibSequence = [1..2000]), iterating through fibSequence linearly X times and for each M item, iterate over offsets linearly N times - 1, meaning offsets reduces by 1 for each iteration :), Ah! Much better!. Giving, M * (N - 1).

The Lesson

I hope the comparison above gives you insight into writing performant code and increasing program performance whenever possible, :), also note the second iteration above, although, performant expects numbers you’re searching through be unique otherwise you’ll end up with an incomplete output.