3차/류동준

3차/류동준_[카카오-인턴]키패드-누르기.py

@@ -0,0 +1,36 @@
+def caldist(l_hand,r_hand, num,position,hand):
+    dist_l = abs(position[num][0] - position[l_hand][0]) + abs(position[num][1] - position[l_hand][1])
+    dist_r = abs(position[num][0] - position[r_hand][0]) + abs(position[num][1] - position[r_hand][1])
+    if dist_l == dist_r:
+        return 'R' if hand =='right' else 'L'
+    return 'R' if dist_l > dist_r else 'L'
+
+def solution(numbers, hand):
+    position = {
+        1: (0,0), 2: (0,1) ,3 :(0,2),
+        4: (1,0), 5: (1,1) ,6 :(1,2),
+        7: (2,0), 8: (2,1) ,9 : (2,2),
+        '*':(3,0), 0:(3,1),'#': (3,2)
+    }
+
+    left_numbers,right_numbers = set([1,4,7,'*']),set([3,6,9,'#'])
+    answer = ''
+    l_hand = '*'
+    r_hand = '#'
+
+    for num in numbers:
+        if num in left_numbers:
+            answer += 'L'
+            l_hand = num
+        elif num in right_numbers:
+            answer += 'R'
+            r_hand = num
+        else:
+            result = caldist(l_hand,r_hand,num,position,hand)
+            if result == 'R':
+                r_hand = num
+            else:
+                l_hand = num
+            answer += result
+
+    return answer

3차/류동준_예산.py

@@ -0,0 +1,17 @@
+def solution(d, budget):
+    answer = 0
+    # d가 가장 작은 값을 순서대로 가지게 끔 함 
+    # 혹은 d의 value를 pop 할때 d.pop(d.index(min(d))) 도 가능하기는함.
+    d.sort()
+    while (budget):
+        # 모든 d 를 pop 하고 budget이 여유있을때도 있음 test case를 통해 확인함.
+        if d:
+            min_value = d.pop(0)
+            if min_value <= budget:
+                budget -= min_value
+                answer += 1
+            else:
+                break
+        else:
+            break
+    return answer

3차/류동준_최대공약수와-최소공배수.py

@@ -0,0 +1,12 @@
+def gcd(a,b):
+    mod = a % b;
+    while(mod > 0):
+        a = b
+        b = mod
+        mod = a % b  
+    return b
+
+
+def solution(n, m):
+    value = gcd(n,m)
+    return [value, n *m / value]

3차/류동준_콜라츠-추축.py

@@ -0,0 +1,13 @@
+def solution(num):
+    answer = 0
+    while(num!=1):
+        if answer > 500:
+            answer = -1
+            break
+        if num % 2 == 0:
+            num = num / 2
+        else:
+            num  = (num * 3) + 1
+        answer += 1
+
+    return answer

3차/류동준_행렬의-덧셈.py

@@ -0,0 +1,7 @@
+# import numpy as np
+def solution(arr1, arr2):
+    for i in range(len(arr1)):
+        for j in range(len(arr1[i])):
+            arr1[i][j] += arr2[i][j]
+    return arr1
+    # return np.add(np.asarray(arr1),np.asarray(arr2)).tolist()