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-rw-r--r--max_diff_remap_digit.c30
-rw-r--r--max_diff_remap_digit.py26
-rw-r--r--min_max_diff_pair.c18
-rw-r--r--min_max_diff_pair.py42
4 files changed, 106 insertions, 10 deletions
diff --git a/max_diff_remap_digit.c b/max_diff_remap_digit.c
new file mode 100644
index 0000000..3eff54a
--- /dev/null
+++ b/max_diff_remap_digit.c
@@ -0,0 +1,30 @@
+// 2566. Maximum Difference by Remapping a Digit
+#include "leetcode.h"
+
+/*
+ * you are given an integer 'num'. you know that bob will sneakily remap one of
+ * the 10 possible digits to another digit. return the difference between the
+ * maximum and minimum values bob can make by remapping exactly one digit in
+ * nums.
+ */
+
+int minMaxDifference(int num) {
+ int min = num, max = num;
+ for (int i = 0; i < 10; ++i) {
+ int low = 0, high = 0, mult = 1;
+ for (int j = num; j; j /= 10) {
+ bool replace = j % 10 == i;
+ low += (replace ? 0 : j % 10) * mult;
+ high += (replace ? 9 : j % 10) * mult;
+ mult *= 10;
+ }
+ min = fmin(min, low);
+ max = fmax(max, high);
+ }
+ return max - min;
+}
+
+int main() {
+ printf("%d\n", minMaxDifference(11891)); // expect: 99009
+ printf("%d\n", minMaxDifference(90)); // expect: 99
+}
diff --git a/max_diff_remap_digit.py b/max_diff_remap_digit.py
new file mode 100644
index 0000000..5d802ed
--- /dev/null
+++ b/max_diff_remap_digit.py
@@ -0,0 +1,26 @@
+# 2566. Maximum Difference by Remapping a Digit
+
+"""
+you are given an integer 'num'. you know that bob will sneakily remap one of
+the 10 possible digits to another digit. return the difference between the
+maximum and minimum values bob can make by remapping exactly one digit in
+nums.
+"""
+
+
+class Solution(object):
+ def minMaxDifference(self, num):
+ """
+ :type num: int
+ :rtype: int
+ """
+ num, i = str(num), 0
+ while num[i] == "9" and i < len(num) - 1:
+ i += 1
+ return int(num.replace(num[i], "9")) - int(num.replace(num[0], "0"))
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minMaxDifference(num=11891))
+ print(obj.minMaxDifference(num=90))
diff --git a/min_max_diff_pair.c b/min_max_diff_pair.c
index 6172bc7..7577e3b 100644
--- a/min_max_diff_pair.c
+++ b/min_max_diff_pair.c
@@ -1,6 +1,5 @@
// 2616. Minimize the Maximum Difference of Pairs
-#include <stdio.h>
-#include <stdlib.h>
+#include "leetcode.h"
/*
* given a 0-indexed integer array 'nums' and an integer 'p'. find 'p' pairs of
@@ -14,13 +13,12 @@
int cmp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
-int minimizeMax(int *nums, int nums_size, int p) {
- qsort(nums, nums_size, sizeof(int), cmp);
- int i, j, left = 0, right = nums[nums_size - 1] - nums[0];
+int minimizeMax(int *nums, int numsSize, int p) {
+ qsort(nums, numsSize, sizeof(int), cmp);
+ int i, j, left = 0, right = nums[numsSize - 1] - nums[0];
while (left < right) {
- int mid = (left + right) / 2;
- int cnt = 0;
- for (int i = 0; i + 1 < nums_size; ++i)
+ int mid = (left + right) / 2, cnt = 0;
+ for (int i = 0; i + 1 < numsSize; ++i)
if (mid >= nums[i + 1] - nums[i])
++cnt, ++i;
if (cnt >= p)
@@ -33,6 +31,6 @@ int minimizeMax(int *nums, int nums_size, int p) {
int main() {
int n1[] = {10, 1, 2, 7, 1, 3}, n2[] = {4, 2, 1, 2};
- printf("%d\n", minimizeMax(n1, 6, 2)); // expect: 1
- printf("%d\n", minimizeMax(n2, 4, 1)); // expect: 0
+ printf("%d\n", minimizeMax(n1, ARRAY_SIZE(n1), 2)); // expect: 1
+ printf("%d\n", minimizeMax(n2, ARRAY_SIZE(n2), 1)); // expect: 0
}
diff --git a/min_max_diff_pair.py b/min_max_diff_pair.py
new file mode 100644
index 0000000..8c5478d
--- /dev/null
+++ b/min_max_diff_pair.py
@@ -0,0 +1,42 @@
+# 2616. Minimize the Maximum Difference of Pairs
+
+"""
+given a 0-indexed integer array 'nums' and an integer 'p'. find 'p' pairs of
+indices of 'nums' such that the maximum difference amongst all the pairs is
+minimised. also, ensure no index appears more than once amongst that 'p'
+pairs. note that for a pair of elements at the index 'i' and 'j'. the
+difference of this pair is '|nums[i] - nums[j]|', where '|x|' represents the
+absolute value of 'x'. return the minimum maximum difference among all 'p'
+pairs. we define the maximum of an empty set to be zero.
+"""
+
+
+class Solution(object):
+ def minimizeMax(self, nums, p):
+ """
+ :type nums: List[int]
+ :type p: int
+ :rtype: int
+ """
+ nums.sort()
+ n = len(nums)
+ left, right = 0, nums[-1] - nums[0]
+ while left < right:
+ mid = (left + right) // 2
+ i, j = 1, 0
+ while i < n:
+ if nums[i] - nums[i - 1] <= mid:
+ i += 1
+ j += 1
+ i += 1
+ if j >= p:
+ right = mid
+ else:
+ left = mid + 1
+ return left
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minimizeMax(nums=[10, 1, 2, 7, 1, 3], p=2))
+ print(obj.minimizeMax(nums=[4, 2, 1, 2], p=1))