[최원준] Day07

leetcode2/1easy/최원준/Q3498.java

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+package Leetcode.최원준;
+
+/*
+1. 아이디어 :
+-
+
+2. 시간복잡도 :
+O( n )
+
+3. 자료구조/알고리즘 :
+- / -
+ */
+
+public class Q3498 {
+    class Solution {
+        public int reverseDegree(String s) {
+            int n = s.length(), ans = 0;
+            for (int i=0; i<n; i++) {
+                ans += (i+1) * (26 - (s.charAt(i) - 'a'));
+            }
+            return ans;
+        }
+    }
+}

leetcode2/2medium/최원준/Q2208.java

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+package Leetcode.최원준;
+
+/*
+1. 아이디어 :
+우선순위큐을 이용해서 가장 큰 값을 반으로 줄이는 작업을 반복
+
+2. 시간복잡도 :
+O( nlogn )
+
+3. 자료구조/알고리즘 :
+우선순위 큐 / -
+ */
+
+import java.util.PriorityQueue;
+
+public class Q2208 {
+    class Solution {
+        public int halveArray(int[] nums) {
+            int n = nums.length, ans = 0;
+            double total = 0.0;
+            PriorityQueue<Double> pq = new PriorityQueue<>((a, b) -> Double.compare(b,a));
+            for (int num : nums) {
+                total += num;
+                pq.add(num + 0.0);
+            }
+
+            double target = total / 2;
+            while (total > target) {
+                ans++;
+                double halfed = pq.poll()/2;
+                total -= halfed;
+                pq.add(halfed);
+            }
+            return ans;
+        }
+    }
+}

leetcode2/3hard/최원준/Q1220.java

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+package Leetcode.최원준;
+
+/*
+1. 아이디어 :
+dp 문제
+
+2. 시간복잡도 :
+O( 5 * n )
+
+3. 자료구조/알고리즘 :
+2차원 배열 / dp
+ */
+
+public class Q1220 {
+    class Solution {
+        int MOD = 1_000_000_007;
+        public int countVowelPermutation(int n) {
+        /*
+        1. e a
+        2. a,i e
+        3. i not i
+        4. i,u o
+        5. u a
+        */
+            long[][] dp = new long[n+1][5];
+            dp[1] = new long[]{1, 1, 1, 1, 1}; // a e i o u
+
+            for (int i=2; i<n+1; i++) {
+                dp[i][0] = (dp[i-1][1]) % MOD;
+                dp[i][1] = (dp[i-1][0] % MOD + dp[i-1][2] % MOD) % MOD;
+                dp[i][2] = (dp[i-1][0] % MOD + dp[i-1][1] % MOD + dp[i-1][3] % MOD + dp[i-1][4] % MOD) % MOD;
+                dp[i][3] = (dp[i-1][2] % MOD + dp[i-1][4] % MOD) % MOD;
+                dp[i][4] = (dp[i-1][0]) % MOD;
+            }
+            long ans = 0;
+            for (int i=0; i<5; i++) ans = (ans + dp[n][i]) % MOD;
+            return (int)ans;
+        }
+    }
+}