• + 2 comments

    Good idea, i did something similar in JavaScript.

    function getMoneySpent (s, n, m, kboards, drives) {
      let kasc = kboards.sort((a,b) => a - b),
        ddesc = drives.sort((a,b) => b - a),
        max = -1,
        sum,
        i = 0,
        j = 0;
    
      do {
        if ((sum = kasc[i] + ddesc[j]) == s)
          return sum;
    
        if (sum < s) {
          if (sum > max)
            max = sum;
          i++;
        } else {
          j++;
        }
      } while (i < n && j < m);
    
      return max;
    }