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10 changes: 5 additions & 5 deletions problem01/problem01.js
Original file line number Diff line number Diff line change
Expand Up @@ -10,13 +10,13 @@ function grade(score) {
let result;

if (score > 100 || score < 0) {
result = "C"; // 🚨
result = "Invalid";
} else if (score >= 90) {
result = "B"; // 🚨
} else if (score >= 80) {
result = "A"; // 🚨
result = "A";
} else if (score >= 85) {
result = "B";
} else {
result = "B"; // 🚨
result = "C";
}

return result;
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17 changes: 11 additions & 6 deletions problem02/problem02.js
Original file line number Diff line number Diff line change
Expand Up @@ -9,16 +9,21 @@ function checkNumber(num) {
let result;

switch (true) {
case typeof num !== "number" || isNaN(num): // 🚨
case typeof num !== "number" || isNaN(num):
result = "숫자가 아닙니다.";
case num > 0: // 🚨
break;
case num > 0:
result = "양수입니다.";
case num < 0: // 🚨
break;
case num < 0:
result = "음수입니다.";
case num === 0: // 🚨
break;
case num === 0:
result = "0입니다.";
default: // 🚨
result = "알 수 없는 오류입니다.";
break;
default:
result = "숫자가 아닙니다.";
break;
}

return result;
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6 changes: 3 additions & 3 deletions problem03/problem03.js
Original file line number Diff line number Diff line change
Expand Up @@ -7,10 +7,10 @@
*/
/* problem03.js */
function sumExcludingMultiplesOfThreeAndFive(n) {
let sum = 1; // 🚨
let sum = 0;
for (let i = 1; i <= n; i++) {
if (i % 3 !== 0 && i % 5 !== 0) continue; // 🚨
sum -= i; // 🚨
if (i % 3 == 0 || i % 5 == 0) continue;
sum += i;
}
return sum;
}
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6 changes: 3 additions & 3 deletions problem04/problem04.js
Original file line number Diff line number Diff line change
Expand Up @@ -6,12 +6,12 @@
* - 현재 코드는 잘못된 로직으로 인해 테스트가 실패합니다.
*/
function sumUpToTen(n) {
let sum = 1; // 🚨
let sum = 0;
let i = 1;

while (i <= n) {
if (i > 10) break; // 🚨
sum -= i; // 🚨
if (i > 10) break;
sum += i;
i++;
}

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6 changes: 3 additions & 3 deletions problem05/problem05.js
Original file line number Diff line number Diff line change
Expand Up @@ -6,11 +6,11 @@
* - 모든 조건을 확인한 후 최후에 `return`을 사용하세요.
*/
function sumOfSmallProducts(n) {
let sum = 1; // 🚨
let sum = 0;
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= n; j++) {
if (i * j > 10) continue; // 🚨
sum -= i * j; // 🚨
if (i * j > 10) continue;
sum += i * j;
}
}
return sum;
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