all-evens.js

/**
 * You can brute force and increment/decrement the given number until you find a valid-all-even-digits
 * number, but the time complexity would be O(n) where n is value of the number.  Instead, you
 * can validate each digit starting with the highest power to come up with the nearest valid number.
 * Since you can make a number even by going up or down, both paths have to be considered.
 * 
 * For example, take 567;
 * 
 * It won't matter what you do with '67' because '5' will always be invalid.  So you have to make it even.
 * 
 * Going up: 5 -> 6     valid!
 * Going down: 5 -> 4   valid!
 * 
 * Then you have to handle the rest, the '67', 
 * 
 * if you are going up: 67 -> 00 because that's the nearest valid higher value i.e. 600
 * if you are going down: 67 -> 88 because that's the nearest valid LOWER value i.e. 488
 * 
 * Now you have the lower and upper, the min of those two from the original number is the minimum clicks
 * the user has to make.
 * 
 * The time complexity of this solution is O(d) where d is the number of digits.
 * 
 */

findMinClicks(42)  		// 0
findMinClicks(2)		// 0
findMinClicks(3)		// 1
findMinClicks(1016) 	// 128
findMinClicks(555)		// 45
findMinClicks(7771)		// 229
findMinClicks(8882) 	// 0

function findMinClicks(number) {
    const upper = findNearestNumberWithEvenDigits(number, +1, '0')
    const lower = findNearestNumberWithEvenDigits(number, -1, '8')
    const min = Math.min(upper - number, number - lower)
    console.log(number, '  upper:', upper, ' lower:', lower, ' -> ', min)
}

function findNearestNumberWithEvenDigits(number, direction, fillDigit) {
	let result = [];
    let num = String(number).split('')
  
    do {
        let [first, ...rest] = num; // first is the highest power digit
    
        if (first % 2 === 0) {      // even - the digit is ok, push it to the result
            result.push(first)
        } else {                    // odd - digit is not ok, increment or decrement to make it even.
            result.push(Number(first) + direction)
            rest.fill(fillDigit)    // fill the rest (insignficant digits) with '0' or '8'
            result = [...result, ...rest] 
            break;                  // we are done, no need to continue;
        }
        
        num = rest
    } while (num.length)
  
    return Number.parseInt(result.join(''))
}