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  1. Practice
  2. Algorithms
  3. Greedy
  4. Luck Balance
  5. Discussions

Luck Balance

  • trkmos + 1 comment

    javaScript solution

    function luckBalance(k, contests) {
    let important = contests.filter(ar => ar[1] === 1).length;
    let sumAll = contests.reduce((a, b) => a+b[0],0);
    let sorted = contests.sort((a, b) => a[0] - b[0])
    let win = important-k >=0 ?important-k : 0 
    let min = 0
    for(let i=0;  i<sorted.length; i++){
        if(win === 0) break;
        if(sorted[i][1] === 0)continue;
        min += sorted[i][0];
        win--
    }
    return sumAll - (2*min);
}
    • jrcastillo2001 + 0 comments

      There seems to be a lot of loops, with this solutions is the time complexity a: O(n*3) + O(m) ... n being constests - m being sorted ... ?

      how about this for a change ?

      function luckBalance(k, contests) {
  const importantContestsVals = []
  let val = 0;
  
  for (let i = 0; i < contests.length; i++) { 
    if (contests[i][1]) {
      importantContestsVals.push(contests[i][0])
    } else { 
      val += contests[i][0]
    }
  }
  
  importantContestsVals.sort((a, b) => a - b)
  const negativeArr = importantContestsVals.splice(0, importantContestsVals.length - k)
  const largestArr = negativeArr.length >= importantContestsVals.length ? negativeArr.length : importantContestsVals.length

  for (let j = 0; j < largestArr; j++) { 
    if (negativeArr[j]) { 
      val -= negativeArr[j]
    }
    if (importantContestsVals[j]) { 
      val += importantContestsVals[j]
    }
  }

  return val
}