skipzero · March 17, 2023 15:06
diff --git a/Sticks.js b/Sticks.js
 /**
 *  in tackling this problem, I simply sorted the array, putting the larger sizes first.
 *  Once I had the array sorted, I started with the biggest size and worked my way through it.
 *  basically just sor tthe array and subtract from the length needed (k)
 * 
 *  I'm not sure if this passes the Big O notation requirement, I assume so with it being
 *  super simple and only having one for loop. That should help keep complexity down.
 * 
 * 
 */

 const getMinSticks = (sticks, k) => {
  let stickCount = 0;
  let remaining = k;
  sticks.sort((a,b) => {
    return a > b ? -1 : 0
  })


  for (let i = 0; i < sticks.length; i++) {
    const length = sticks[i];
    const needed = Math.floor(remaining / length);
    let used = Math.min(needed, sticks[i])
    stickCount += used;
    remaining -= used * length;
  }
  console.log('returned', remaining === 0 ? stickCount : -1)
  return remaining === 0 ? stickCount : -1
 }

 const sticks = [1, 2, 3, 4];
 const k = 8;

 getMinSticks(sticks, 25)
	/**
	* in tackling this problem, I simply sorted the array, putting the larger sizes first.
	* Once I had the array sorted, I started with the biggest size and worked my way through it.
	* basically just sor tthe array and subtract from the length needed (k)
	*
	* I'm not sure if this passes the Big O notation requirement, I assume so with it being
	* super simple and only having one for loop. That should help keep complexity down.
	*
	*
	*/

	const getMinSticks = (sticks, k) => {
	let stickCount = 0;
	let remaining = k;
	sticks.sort((a,b) => {
	return a > b ? -1 : 0
	})


	for (let i = 0; i < sticks.length; i++) {
	const length = sticks[i];
	const needed = Math.floor(remaining / length);
	let used = Math.min(needed, sticks[i])
	stickCount += used;
	remaining -= used * length;
	}
	console.log('returned', remaining === 0 ? stickCount : -1)
	return remaining === 0 ? stickCount : -1
	}

	const sticks = [1, 2, 3, 4];
	const k = 8;

	getMinSticks(sticks, 25)