daily

Author: Zanger67

my-submissions/e3461 v3 combinatorics O(n).py

@@ -3,7 +3,7 @@ def hasSameDigits(self, s: str) -> bool:
         a, b = 0, 0
         x = len(s) - 2
         for i, num in enumerate(s[:-1]) :
-            a += (int(num) * comb(x, i)) % 100
+            a += (int(num) * comb(x, i)) % 10
         for i, num in enumerate(s[1:]) :
-            b += (int(num) * comb(x, i)) % 100
+            b += (int(num) * comb(x, i)) % 10
         return a % 10 == b % 10

my-submissions/e3461 v4.py

@@ -0,0 +1,6 @@
+class Solution:
+    def hasSameDigits(self, s: str) -> bool:
+        x = len(s) - 2
+        a = sum((int(num) * comb(x, i)) % 10 for i, num in enumerate(s[:-1])) % 10
+        b = sum((int(num) * comb(x, i)) % 10 for i, num in enumerate(s[1:])) % 10
+        return a == b

my-submissions/e3461.md

@@ -0,0 +1,60 @@
+## Notes
+
+### V1/V2
+
+Simulation method -- inefficient
+
+### V3 and above
+
+Combinatorics to create an $O(n)$ solution.
+
+Breaking down examples
+
+```
+s="12345"
+
+1 2 3 4 5
+3 5 7 9
+(1+2)mod10 (2+3)mod10 (3+4)mod10 (4+5)mod10
+
+8 12 16
+8 2 6
+
+((1+2)mod10+(2+3)mod10)mod10 ((2+3)mod10+(3+4)mod10)mod10
+((3+4)mod10+(4+5)mod10)mod10
+
+(1+2+2+3)mod10 (2+3+3+4)mod10
+(3+4+4+5)mod10
+
+(1+2+2+2+3+3+3+4)mod10 (2+3+3+3+4+4+4+5)mod10
+
+(1+2+2+2+3+3+3+4+2+3+3+3+4+4+4+5)mod10
+(1+2+2+2+2+3+3+3+3+3+3+4+4+4+4+5)mod10
+(1*1+4*2+6*3+3*4+1*5)mod10 = 44 => TRUE
+(1*a[0]+4*a[1]+6*a[2]+3*a[3]+1*a[4])mod10
+
+(1*a[0]+4*a[1]+6*a[2]+3*a[3]+1\*a[4])mod10
+
+This works for this case of 12345
+
+3902
+3 9 0 2
+12 9 2
+2 9 2
+
+11 11
+1 1 => TRUE
+
+3 9 0 2
+(3+9)mod10 (9+0)mod10 (0+2)mod10
+(3+9+9+0)mod10 (9+0+0+2)mod10 = 21mod10 11mod10 = 1 1 => TRUE => CORRECT
+(1*a[0]+2*a[1]+1\*a[2])
+
+(3+9+9+0+9+0+0+2)mod10
+= 32 => FALSE => WRONG
+
+We find an edge case if we go all the way. However, if we stop one step before... we can compare their digit vals.
+
+Seems like we should stop a step before and perform our comparison?
+2 operations of O(n)
+```

my-submissions/h3463.py

@@ -0,0 +1,6 @@
+class Solution:
+    def hasSameDigits(self, s: str) -> bool:
+        x = len(s) - 2
+        a = sum((int(num) * comb(x, i)) % 10 for i, num in enumerate(s[:-1])) % 10
+        b = sum((int(num) * comb(x, i)) % 10 for i, num in enumerate(s[1:])) % 10
+        return a == b

my-submissions/m2048 v1 brute force.py

@@ -0,0 +1,7 @@
+class Solution:
+    def nextBeautifulNumber(self, n: int) -> int:
+        is_beaut = lambda x: all(int(k) == v for k, v in Counter(str(x)).items())
+        n += 1
+        while not is_beaut(n) :
+            n += 1
+        return n