Update Uniform_Integers.py

Level 1/Uniform_Integers.py

@@ -9,76 +9,25 @@
 ## 1 <= A <= B <= 10^(12)
 
 ## Solution
-## Time Complexity: O(log10(A * B)) == O(log10(A) + log10(B))
-## Space Complexity: O(log10(A * B))
-## Explanation: Given an uniform integer D, we can determine how many uniform
-## integers are less than or equal to D in O(log10(D)) time. So to find the
-## number of uniform integers between A and B, we just need to subtract the
-## number of uniform integers less than B from the number of uniform integers
-## less than A. To find the smallest uniform integer greater than A, we
-## need to do is make a uniform integer by changing all of A's digits to A's
-## first digit. For example, 2468753 becomes 2222222. If it is smaller than A
-## then the next uniform integer will be greater than A. This takes O(log10(A))
-## time for A and O(log10(B)) for B, resulting in O(log10(A) + log10(B)) time.
-
-## Here's the pattern:
-## 0 * 9 + 1 -> 1  |  1 * 9 + 1 -> 11  |  2 * 9 + 1 -> 111  |  3 * 9 + 1 -> 1111
-## 0 * 9 + 2 -> 2  |  1 * 9 + 2 -> 22  |  2 * 9 + 2 -> 222  |  3 * 9 + 2 -> 2222
-## 0 * 9 + 3 -> 3  |  1 * 9 + 3 -> 33  |  2 * 9 + 3 -> 333  |  3 * 9 + 3 -> 3333
-## 0 * 9 + 4 -> 4  |  1 * 9 + 4 -> 44  |  2 * 9 + 4 -> 444  |  3 * 9 + 4 -> 4444
-## 0 * 9 + 5 -> 5  |  1 * 9 + 5 -> 55  |  2 * 9 + 5 -> 555  |  3 * 9 + 5 -> 5555
-## 0 * 9 + 6 -> 6  |  1 * 9 + 6 -> 66  |  2 * 9 + 6 -> 666  |  3 * 9 + 6 -> 6666
-## 0 * 9 + 7 -> 7  |  1 * 9 + 7 -> 77  |  2 * 9 + 7 -> 777  |  3 * 9 + 7 -> 7777
-## 0 * 9 + 8 -> 8  |  1 * 9 + 8 -> 88  |  2 * 9 + 8 -> 888  |  3 * 9 + 8 -> 8888
-## 0 * 9 + 9 -> 9  |  1 * 9 + 9 -> 99  |  2 * 9 + 9 -> 999  |  3 * 9 + 9 -> 9999
-
+## Time Complexity: O(9 * log10(B))
+## Space Complexity: O(1)
+## Explanation:
+'''
+    n = [1..9]
+                         v0 = 0
+    1   = 1 + 0  + 0     v1 = v0 * 10 + n = n  shift v to left and add n
+    11  = 1 + 10 + 0     v2 = v1 * 10 + n
+    111 = 1 + 10 + 100   v3 = v2 * 10 + n
+'''
 def getUniformIntegerCountInInterval(A: int, B: int) -> int:
-    first_uniform_int_id = 0
-    last_uniform_int_id  = 0
-
-    ## Convert integer to list of digits
-    integer_digits = []
-    while A > 0:
-        integer_digits.append(A % 10)
-        A = A // 10
-
-    ## Loop through digits from left to right
-    first_digit = integer_digits[-1]
-    for i in range(len(integer_digits) - 2, -1, -1):
-        ## The uniform integer is greater than A
-        if integer_digits[i] < first_digit:
-            break
-
-        ## The uniform integer is less than A 
-        if integer_digits[i] > first_digit:
-            first_digit += 1
-            break
-
-    ## Map uniform integer to its counting ID
-    first_uniform_int_id = (9 * (len(integer_digits) - 1)) + first_digit
-
-    ## Convert integer to list of digits
-    integer_digits = []
-    while B > 0:
-        integer_digits.append(B % 10)
-        B = B // 10
-
-    ## Loop through digits from left to right
-    first_digit = integer_digits[-1]
-    for i in range(len(integer_digits) - 2, -1, -1):
-        ## The uniform integer is less than B 
-        if integer_digits[i] > first_digit:
-            break
-
-        ## The uniform integer is greater than B
-        if integer_digits[i] < first_digit:
-            first_digit -= 1
-            break
-
-    ## Map uniform integer to its counting ID
-    last_uniform_int_id = (9 * (len(integer_digits) - 1)) + first_digit
-
-    return last_uniform_int_id - first_uniform_int_id + 1
+  total = 0
+  for i in range(1, 10): # O(9)
+    v = i
+    while v <= B:        # O(log(B))
+      if A <= v <=B:
+        total += 1
+      v = v * 10 + i
+  return total
 
 ## Test Cases
 if __name__ == "__main__":