[khyo] Week 2

3sum/higeuni.js

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+// complexity
+// time: O(n^2)
+// - sort: O(n log n)
+// - for loop: O(n)
+// - while loop: O(n)
+// space: O(n)
+// - sortedNums: O(n)
+// - else : O(1)
+
+var threeSum = function(nums) {
+  const sortedNums = nums.sort((a, b) => a - b)
+  const lengthOfArray = nums.length;
+  const answer = [];
+
+  for(let i = 0; i < lengthOfArray; i++ ) {
+      if (i > 0 && nums[i] === nums[i - 1]) continue;
+      const target = (-1) * sortedNums[i];
+
+      let left = i + 1;
+      let right = lengthOfArray - 1;
+
+      while(left < right) {
+          const sumOfLeftAndRight = sortedNums[left] + sortedNums[right];
+          const diff = sumOfLeftAndRight - target;
+
+          if( diff > 0 ){
+              right -= 1;
+          } else if ( diff < 0 ){
+              left += 1
+          } else {
+              answer.push([sortedNums[i], sortedNums[left], sortedNums[right]]);
+              // 중복되는 부분을 처리하는 과정에서 계속 fail되어 찾아보니 이렇게 해야 통과되었다.
+              while(left < right && sortedNums[left] === sortedNums[left + 1]) left ++;
+              while(left < right && sortedNums[right] === sortedNums[right - 1]) right --;
+
+              left++;
+              right--;
+          }
+      }
+  }
+  return answer
+};
+

climbing-stairs/higeuni.js

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+// complexity
+// time: O(n)
+// - for loop: O(n)
+// space: O(1)
+// - n에 관계없이 상수개의 변수만 사용되므로 O(1)
+
+var climbStairs = function(n) {
+  let num1 = 1;
+  let num2 = 1;
+  let temp = 0;
+
+  for(let i = 2; i<=n; ++i ) {
+      temp = num2;
+      num2 = num1 + num2;
+      num1 = temp;
+  }
+
+  return num2;
+};
+

valid-anagram/higeuni.js

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+// complexity
+// time: O(n log n)
+// - sort: O(n log n)
+// - split: O(n)
+// - join: O(n)
+// space: O(n)
+// - sortedS: O(n)
+// - sortedT: O(n)
+// - else : O(1)
+
+var isAnagram = function(s, t) {
+  return s.split('').sort().join('') === t.split('').sort().join('')
+};
+