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-rw-r--r--lex_smallest_beaut_str.c47
-rw-r--r--lex_smallest_beaut_str.py36
-rw-r--r--max_diff_between_adj_elem.c23
-rw-r--r--max_diff_between_adj_elem.py27
-rw-r--r--max_diff_even_odd_freq2.c120
-rw-r--r--max_diff_even_odd_freq2.py78
-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
10 files changed, 437 insertions, 10 deletions
diff --git a/lex_smallest_beaut_str.c b/lex_smallest_beaut_str.c
new file mode 100644
index 0000000..8009a4a
--- /dev/null
+++ b/lex_smallest_beaut_str.c
@@ -0,0 +1,47 @@
+// 2663. Lexicographically Smallest Beautiful String
+#include "leetcode.h"
+
+/*
+ * a string is beautiful if it consists of the first 'k' letters of the english
+ * lowercase alphabet, it does not contain a palindrome of length 2 or more
+ * which is a palindrome. you are given a beautiful string 's' of length 'n',
+ * and a positive integer 'k'. return the lexicographically smallest string of
+ * length 'n' which is larger than 's', and is beautiful. if there is no such
+ * string, return an empty string.
+ */
+
+char *smallestBeautifulString(char *s, int k) {
+ int n = strlen(s);
+ char *ans = (char *)malloc((n + 1) * sizeof(char));
+ strcpy(ans, s);
+ for (int i = n - 1; i >= 0; i--) {
+ for (int j = ans[i] - 'a' + 1; j < k; j++) {
+ char curr = j + 'a';
+ if ((i < 1 || curr != ans[i - 1]) && (i < 2 || curr != ans[i - 2])) {
+ ans[i] = curr;
+ for (int l = i + 1; l < n; l++) {
+ for (int m = 0; m < k; m++) {
+ curr = m + 'a';
+ if ((l < 1 || curr != ans[l - 1]) &&
+ (l < 2 || curr != ans[l - 2])) {
+ ans[l] = curr;
+ break;
+ }
+ }
+ }
+ return ans;
+ }
+ }
+ }
+ return "";
+}
+
+int main() {
+ char *s1 = "abcz", *s2 = "dc";
+ char *sbs1 = smallestBeautifulString(s1, 26);
+ char *sbs2 = smallestBeautifulString(s2, 4);
+ printf("%s\n", sbs1); // expect: abda
+ printf("%s\n", sbs2); // expect: null
+ free(sbs1);
+ free(sbs2);
+}
diff --git a/lex_smallest_beaut_str.py b/lex_smallest_beaut_str.py
new file mode 100644
index 0000000..58e1381
--- /dev/null
+++ b/lex_smallest_beaut_str.py
@@ -0,0 +1,36 @@
+# 2663. Lexicographically Smallest Beautiful String
+
+"""
+a string is beautiful if it consists of the first 'k' letters of the english
+lowercase alphabet, it does not contain a palindrome of length 2 or more
+which is a palindrome. you are given a beautiful string 's' of length 'n',
+and a positive integer 'k'. return the lexicographically smallest string of
+length 'n' which is larger than 's', and is beautiful. if there is no such
+string, return an empty string.
+"""
+
+
+class Solution(object):
+ def smallestBeautifulString(self, s, k):
+ """
+ :type s: str
+ :type k: int
+ :rtype: str
+ """
+ ans = [ord(c) - ord("a") for c in s]
+ n = len(ans) - 1
+ while n >= 0:
+ ans[n] += 1
+ if ans[n] == k:
+ n -= 1
+ elif ans[n] not in ans[max(n - 2, 0) : n]:
+ for j in range(n + 1, len(ans)):
+ ans[j] = min({0, 1, 2} - set(ans[max(0, j - 2) : j]))
+ return "".join(chr(ord("a") + i) for i in ans)
+ return ""
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.smallestBeautifulString(s="abcz", k=26))
+ print(obj.smallestBeautifulString(s="dc", k=4))
diff --git a/max_diff_between_adj_elem.c b/max_diff_between_adj_elem.c
new file mode 100644
index 0000000..a96f02c
--- /dev/null
+++ b/max_diff_between_adj_elem.c
@@ -0,0 +1,23 @@
+// 3423. Maximum Difference Between Adjacent Elements in a Circular Array
+#include "leetcode.h"
+
+/*
+ * given a circular array 'nums', find the maximum absolute difference between
+ * adjacent elements. note, in a circular array, the first and last elements are
+ * adjacent.
+ */
+
+int maxAdjacentDistance(int *nums, int numsSize) {
+ int ans = abs(nums[0] - nums[numsSize - 1]);
+ for (int i = 1; i < numsSize; i++) {
+ int curr = abs(nums[i - 1] - nums[i]);
+ ans = fmax(ans, curr);
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {1, 2, 4}, n2[] = {-5, -10, -5};
+ printf("%d\n", maxAdjacentDistance(n1, ARRAY_SIZE(n2))); // expect: 3
+ printf("%d\n", maxAdjacentDistance(n1, ARRAY_SIZE(n2))); // expect: 5
+}
diff --git a/max_diff_between_adj_elem.py b/max_diff_between_adj_elem.py
new file mode 100644
index 0000000..4334537
--- /dev/null
+++ b/max_diff_between_adj_elem.py
@@ -0,0 +1,27 @@
+# 3423. Maximum Difference Between Adjacent Elements in a Circular Array
+
+"""
+given a circular array 'nums', find the maximum absolute difference between
+adjacent elements. note, in a circular array, the first and last elements are
+adjacent.
+"""
+
+
+class Solution(object):
+ def maxAdjacentDistance(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ n = len(nums)
+ ans = abs(nums[0] - nums[n - 1])
+ for i in range(1, n):
+ curr = abs(nums[i - 1] - nums[i])
+ ans = max(ans, curr)
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxAdjacentDistance(nums=[1, 2, 4]))
+ print(obj.maxAdjacentDistance(nums=[-5, -10, -5]))
diff --git a/max_diff_even_odd_freq2.c b/max_diff_even_odd_freq2.c
new file mode 100644
index 0000000..8d32226
--- /dev/null
+++ b/max_diff_even_odd_freq2.c
@@ -0,0 +1,120 @@
+// 3445. Maximum Difference Between Even and Odd Frequency II
+#include "leetcode.h"
+
+/*
+ * you are given a string 's' and an integer 'k'. your task is to find the
+ * maximum difference between the frequency of two characters 'freq[a], -
+ * freq[b]', in a substring 'subs' of 's' such that 'subs' has a size of at
+ * least 'k', character 'a' has an odd frequency in 'subs', character 'b' has an
+ * even frequency in 'subs'. return the maximum difference. note that 'subs' can
+ * contain more than 2 distinct characters.
+ */
+
+static int upper_bound(const int *arr, int n, int key) {
+ int low = 0, high = n;
+ while (low < high) {
+ int mid = low + ((high - low) >> 1);
+ if (arr[mid] <= key)
+ low = mid + 1;
+ else
+ high = mid;
+ }
+ return low;
+}
+
+int maxDifference(char *s, int k) {
+ int n = strlen(s);
+ if (!n || k > n)
+ return -1;
+ int (*freq)[5] = malloc(sizeof(int[5]) * n), present[5] = {0};
+ if (!freq)
+ return -1;
+ for (int i = 0; i < n; i++) {
+ int c = s[i] - '0';
+ if (!i)
+ memset(freq[i], 0, sizeof(int) * 5);
+ else
+ memcpy(freq[i], freq[i - 1], sizeof(int) * 5);
+ freq[i][c]++;
+ present[c] = 1;
+ }
+ int ans = INT_MIN;
+ for (int a = 0; a < 5; a++) {
+ if (!present[a])
+ continue;
+ for (int b = 0; b < 5; b++) {
+ if (!present[b] || a == b)
+ continue;
+ int *idx_list[4], *val_list[4], size[4];
+ for (int i = 0; i < 4; i++) {
+ idx_list[i] = (int *)malloc(sizeof(int) * n);
+ val_list[i] = (int *)malloc(sizeof(int) * n);
+ size[i] = 0;
+ }
+ int odd = 0, even = 0;
+ for (int i = 0; i < n; i++) {
+ int c = s[i] - '0';
+ if (c == a)
+ odd++;
+ else if (c == b)
+ even++;
+ if ((odd & 1) && (even & 1) == 0 && (i + 1) >= k && even > 0) {
+ int diff = odd - even;
+ if (diff > ans)
+ ans = diff;
+ }
+ int key1 = ((odd % 2 == 0) ? 1 : 0) << 1 | ((even % 2 == 0) ? 0 : 1);
+ if (size[key1] > 0) {
+ int limit = i - k;
+ if (limit >= 0) {
+ int pos = upper_bound(idx_list[key1], size[key1], limit);
+ if (pos > 0) {
+ int prev = pos - 1, l = 0, r = prev, curr_min = INT_MAX;
+ while (l <= r) {
+ int mid = (l + r) >> 1, j = idx_list[key1][mid];
+ if (freq[i][b] != freq[j][b]) {
+ int v = val_list[key1][mid];
+ if (v < curr_min)
+ curr_min = v;
+ l = mid + 1;
+ } else
+ r = mid - 1;
+ }
+ if (curr_min != INT_MAX) {
+ int cur = odd - even - curr_min;
+ if (cur > ans)
+ ans = cur;
+ }
+ }
+ }
+ }
+ int key2 = (odd % 2) << 1 | (even % 2), curr_diff = odd - even;
+ if (!size[key2]) {
+ idx_list[key2][0] = i;
+ val_list[key2][0] = curr_diff;
+ size[key2] = 1;
+ } else {
+ int prev_min = val_list[key2][size[key2] - 1];
+ int new_min = (prev_min < curr_diff) ? prev_min : curr_diff;
+ idx_list[key2][size[key2]] = i;
+ val_list[key2][size[key2]] = new_min;
+ size[key2]++;
+ }
+ }
+ for (int i = 0; i < 4; i++) {
+ free(idx_list[i]);
+ free(val_list[i]);
+ }
+ }
+ }
+
+ free(freq);
+ return (ans == INT_MIN ? -1 : ans);
+}
+
+int main() {
+ char *s1 = "12233", *s2 = "1122211", *s3 = "110";
+ printf("%d\n", maxDifference(s1, 4)); // expect: -1
+ printf("%d\n", maxDifference(s2, 3)); // expect: 1
+ printf("%d\n", maxDifference(s3, 3)); // expect: -1
+}
diff --git a/max_diff_even_odd_freq2.py b/max_diff_even_odd_freq2.py
new file mode 100644
index 0000000..a24d3d0
--- /dev/null
+++ b/max_diff_even_odd_freq2.py
@@ -0,0 +1,78 @@
+# 3445. Maximum Difference Between Even and Odd Frequency II
+
+"""
+you are given a string 's' and an integer 'k'. your task is to find the
+maximum difference between the frequency of two characters 'freq[a], -
+freq[b]', in a substring 'subs' of 's' such that 'subs' has a size of at
+least 'k', character 'a' has an odd frequency in 'subs', character 'b' has an
+even frequency in 'subs'. return the maximum difference. note that 'subs' can
+contain more than 2 distinct characters.
+"""
+
+
+class fenwick_tree:
+ inf = float("inf")
+
+ def __init__(self, n):
+ self.n = n
+ self.tree = [self.inf] * (n + 1)
+
+ def update(self, i, val):
+ pos = i + 1
+ while pos <= self.n:
+ self.tree[pos] = min(self.tree[pos], val)
+ pos += pos & -pos
+
+ def query(self, i):
+ res, pos = self.inf, i + 1
+ while pos > 0:
+ res = min(res, self.tree[pos])
+ pos -= pos & -pos
+ return res
+
+
+class Solution(object):
+ def maxDifference(self, s, k):
+ """
+ :type s: str
+ :type k: int
+ :rtype: int
+ """
+ n, ans = len(s), float("-inf")
+ for a in range(5):
+ for b in range(5):
+ if a == b:
+ continue
+ d = [0] * (n + 1)
+ pa, pb = [0] * (n + 1), [0] * (n + 1)
+ cnt_b = [0] * (n + 1)
+ for i in range(1, n + 1):
+ digit = int(s[i - 1])
+ d[i] = (
+ d[i - 1] + (1 if digit == a else 0) - (1 if digit == b else 0)
+ )
+ pa[i] = (pa[i - 1] + (1 if digit == a else 0)) & 1
+ pb[i] = (pb[i - 1] + (1 if digit == b else 0)) & 1
+ cnt_b[i] = cnt_b[i - 1] + (1 if digit == b else 0)
+ trees = [[fenwick_tree(n + 1) for _ in range(2)] for _ in range(2)]
+ for j in range(n + 1):
+ if j >= k:
+ idx = j - k
+ trees[pa[idx]][pb[idx]].update(cnt_b[idx], d[idx])
+ if j > 0:
+ need_p, need_q = (
+ 1 - pa[j],
+ pb[j],
+ )
+ if cnt_b[j] > 0:
+ best = trees[need_p][need_q].query(cnt_b[j] - 1)
+ if best != fenwick_tree.inf:
+ ans = max(ans, d[j] - best)
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxDifference(s="12233", k=4))
+ print(obj.maxDifference(s="1122211", k=3))
+ print(obj.maxDifference(s="110", k=3))
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))