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[Sehwan]Added solutions for week1 #13
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,30 @@ | ||
/** | ||
* @param {number[]} prices | ||
* @return {number} | ||
*/ | ||
var maxProfit = function (prices) { | ||
// Initiate left and right pointer (left: buy price / right: sell price) | ||
// Make profit with initiative value 0 | ||
let l = 0; | ||
let r = 1; | ||
let profit = 0; | ||
// Iterate to check profit = prices[r] - prices[l] | ||
while (r < prices.length) { | ||
// If profit is positive, compare profit with previous one | ||
if (prices[r] > prices[l]) { | ||
profit = Math.max(profit, prices[r] - prices[l]); | ||
r++; | ||
} else { | ||
// If profit is negative, move forware left and right pointer | ||
l = r; | ||
r++; | ||
} | ||
} | ||
return profit; | ||
}; | ||
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// TC: O(n) | ||
// SC: O(3) | ||
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console.log(maxProfit([7, 1, 5, 3, 6, 4])); //5 | ||
console.log(maxProfit([7, 6, 4, 3, 1])); //0 |
Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,24 @@ | ||
/** | ||
* @param {number[]} nums | ||
* @return {boolean} | ||
*/ | ||
var containsDuplicate = function (nums) { | ||
// 1. Make hashmap that has previous value | ||
let map = {}; | ||
// 2. Iterate nums to check nums[i] is duplicated or not | ||
for (let i = 0; i < nums.length; i++) { | ||
// 2.1 Check there is same value on the map | ||
// 3. Return true when their is duplicated value | ||
if (nums[i] in map) { | ||
return true; | ||
} else map[nums[i]] = i; | ||
} | ||
return false; | ||
}; | ||
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// TC: O(n) | ||
// SC: O(n) | ||
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console.log(containsDuplicate([1, 2, 3, 1])); // true | ||
console.log(containsDuplicate([1, 2, 3, 4])); // false | ||
console.log(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])); // true | ||
Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,23 @@ | ||
/** | ||
* @param {number[]} nums | ||
* @param {number} target | ||
* @return {number[]} | ||
*/ | ||
var twoSum = function (nums, target) { | ||
// 1. Make a Hashmap to store { key(index) : value(target-num)} | ||
let map = {}; | ||
// 2. Iterate to find value is equal to (target - nums[i]) | ||
// There is only one solution | ||
for (let i = 0; i < nums.length; i++) { | ||
const diff = target - nums[i]; | ||
// 3. If there is an index that has different value, return array | ||
if (diff in map) return [map[diff], i]; | ||
map[nums[i]] = i; | ||
} | ||
}; | ||
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// TC: O(n) | ||
// SC: O(n) | ||
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console.log(twoSum([2, 7, 11, 15], 9)); | ||
console.log(twoSum([3, 2, 4], 6)); |
Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,38 @@ | ||
/** | ||
* @param {string} s | ||
* @param {string} t | ||
* @return {boolean} | ||
*/ | ||
var isAnagram = function (s, t) { | ||
/* | ||
// Split string and sort charaters | ||
// Compare s and t after sorting | ||
return s.split("").sort().join() === t.split("").sort().join() | ||
// 92ms, 53.29MB | ||
*/ | ||
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// Made Hashmap and count number of each charater | ||
let map = {}; | ||
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// Check character length between s and t | ||
if (s.length !== t.length) return false; | ||
// Put a character and add or substract number | ||
for (let i = 0; i < s.length; i++) { | ||
// map[s[i]] = map[s[i]] ? map[s[i]] + 1 : 1; | ||
map[s[i]] = (map[s[i]] ?? 0) + 1; | ||
// map[t[i]] = map[t[i]] ? map[t[i]] - 1 : -1; | ||
map[t[i]] = (map[t[i]] ?? 0) - 1; | ||
} | ||
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for (let i = 0; i < s.length; i++) { | ||
if (map[s[i]] !== 0) return false; | ||
} | ||
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return true; | ||
}; | ||
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console.log(isAnagram("anagram", "nagaram")); | ||
console.log(isAnagram("rat", "car")); | ||
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// TC: O(n) | ||
// SC: O(n) |
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@@ -0,0 +1,43 @@ | ||
/** | ||
* @param {string} s | ||
* @return {boolean} | ||
*/ | ||
var isPalindrome = function (s) { | ||
// 1. Create left and right pointer | ||
let l = 0; | ||
let r = s.length - 1; | ||
// 2. Iterate while loop with l<r condition | ||
while (l < r) { | ||
// 3. Check character with left pointer is alphanumeric character or not | ||
if ( | ||
(s[l] >= "a" && s[l] <= "z") || | ||
(s[l] >= "A" && s[l] <= "Z") || | ||
(s[l] >= "0" && s[l] <= "9") | ||
) { | ||
// 4. Check character with right pointer is alphanumeric character or not | ||
if ( | ||
(s[r] >= "a" && s[r] <= "z") || | ||
(s[r] >= "A" && s[r] <= "Z") || | ||
(s[r] >= "0" && s[r] <= "9") | ||
) { | ||
// 5. Compare left and right pointer character | ||
if (s[l].toLowerCase() !== s[r].toLowerCase()) { | ||
return false; | ||
} else { | ||
l++; | ||
r--; | ||
} | ||
// If not, go to next location for right pointer | ||
} else r--; | ||
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// If not, go to next location for left pointer | ||
} else l++; | ||
Comment on lines
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. @nhistory 님, 코드가 이미 머지가 되었지만 사소한 피드백을 드리고 싶어서 코멘트 드립니다. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. @DaleSeo 귀한 피드백 감사드립니다! // 2. Iterate while loop with l<r condition
while (l < r) {
// 3. Check left and right pointer has alphanumeric character
while (l < r && !isAlphanumeric(s[l])) l++;
while (l < r && !isAlphanumeric(s[r])) r--;
// 4. Compare left and right pointer character
if (s[l].toLowerCase() !== s[r].toLowerCase()) {
return false;
}
l++;
r--;
}
return true;
}; |
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} | ||
return true; | ||
}; | ||
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// TC: O(n) | ||
// SC: O(n) | ||
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console.log(isPalindrome("A man, a plan, a canal: Panama")); //true | ||
console.log(isPalindrome("race a car")); //false |
There was a problem hiding this comment.
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The reason will be displayed to describe this comment to others. Learn more.
캬~ 복잡도 분석에 셀프 테스트까지? 멋지십니다! 👏👏👏