Leetcode contest
Q3. Minimum Operations to Transform Array Status: Medium #leetcodecontest #leetcode #contest
i will share q3 after 510 subcriber
Leetcode contest
Q3. Minimum Operations to Transform Array Status: Medium #leetcodecontest #leetcode #contest
Problem:Q3. β
Status: Hardπ΄
Code: Kotlinπ
@leetcode7 @contest #leetcode #biweekly
Status: Hard
Code: Kotlin
class Solution {
fun minOperations(nums1: IntArray, nums2: IntArray): Long {
val travenior = nums1.toMutableList()
val n = nums1.size
val x = nums2[n].toLong()
var baseSum = 0L
var bestExtra = Long.MAX_VALUE
for (i in 0 until n) {
val a = nums1[i].toLong()
val b = nums2[i].toLong()
baseSum += kotlin.math.abs(a - b)
val l = kotlin.math.min(a, b)
val r = kotlin.math.max(a, b)
val dist = when {
x < l -> l - x
x > r -> x - r
else -> 0L
}
val extra = dist + 1L
if (extra < bestExtra) bestExtra = extra
}
return baseSum + bestExtra
}
}@leetcode7 @contest #leetcode #biweekly
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Forwarded from SeedOSS β’ foss, privacy, gems.
Sozo TV β Android TV
β¨ Why it's amazing
π Get it: Telegram
π² Screenshots
π Source Code
β€οΈ Sincerely: @SeedOSS
Sozo is crafted with a perfect blend of simplicity and state-of-the-art elegance. It is designed as an Anilist only client, offering users the ability to stream and download their favorite Anime, Manga, Movies, and TV Shows.
β¨ Why it's amazing
π Get it: Telegram
π² Screenshots
π Source Code
β€οΈ Sincerely: @SeedOSS
Farmers who wait for the perfect weather never plannt.
βββββββββββ
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Forwarded from Azamov | Dev
Perhaps the thing I'm most worried about in the future of AI and autofill is that I don't want to see ads in code either
#ai
#
Forwarded from π½πΌπ½π π’ππ¦ | ππ’π¦π¦, ππΆπ³π², π π²πΊπ²π (κ πππππππππ ལསΰΌΰ½ ΰ½ΰ½’)
Sozo for AndroidTV
Sozo is crafted with a perfect blend of simplicity and state-of-the-art elegance. It is designed as an Anilist only client, offering users the ability to stream and download their favorite Anime & Manga for AndroidTV too
π Links:
- Download
- Screenshots
- Support
- Source code
Developer: ProfessorDeveloper
β οΈ This app might be breaking copyright laws, use at your own risk. We are not responsible for anything. We are just sharing open source software.
π· Tags: #Anime #Movie #AndroidTV
Sozo is crafted with a perfect blend of simplicity and state-of-the-art elegance. It is designed as an Anilist only client, offering users the ability to stream and download their favorite Anime & Manga for AndroidTV too
π Links:
- Download
- Screenshots
- Support
- Source code
Developer: ProfessorDeveloper
β€οΈ Support the Project
If this project makes your life easier, here are a few quick ways to show some love:
β Star the repo/app
β Buy a coffee for the developer
π Contribute code, issues, or pull-requests
β οΈ This app might be breaking copyright laws, use at your own risk. We are not responsible for anything. We are just sharing open source software.
π· Tags: #Anime #Movie #AndroidTV
β€3
class Solution {
public int minMoves(int[] nums) {
int max = nums[0];
long sum = 0;
for (int x : nums) {
if (x > max) max = x;
sum += x;
}
return (int)((long) max * nums.length - sum);
}
}
@leetcode7
class Solution {
public int countMajoritySubarrays(int[] nums, int target) {
int n = nums.length;
int[] pref = new int[n + 1]; // pref[0] = 0 by default
int[] dresaniel = nums.clone();
for (int i = 0; i < n; i++) {
pref[i + 1] = pref[i] + (dresaniel[i] == target ? 1 : -1);
}
int ans = 0;
for (int r = 1; r <= n; r++) {
for (int l = 0; l < r; l++) {
if (pref[r] - pref[l] > 0) ans++;
}
}
return ans;
}
}
Β©leetcode7@leetcode7
Q2
Forwarded from Azamov
Q3. Longest Non-Decreasing Subarray After Replacing at Most One Element
Code: Java
Code: Java
class Solution {
public int longestSubarray(int[] nums) {
int n = nums.length;
if (n <= 1) return n;
int[] left = new int[n];
left[0] = 1;
for (int i = 1; i < n; i++) {
left[i] = (nums[i] >= nums[i - 1]) ? left[i - 1] + 1 : 1;
}
int[] serathion = new int[n];
System.arraycopy(nums, 0, serathion, 0, n);
int[] right = new int[n];
right[n - 1] = 1;
for (int i = n - 2; i >= 0; i--) {
right[i] = (serathion[i] <= serathion[i + 1]) ? right[i + 1] + 1 : 1;
}
int ans = 0;
for (int i = 0; i < n; i++) ans = Math.max(ans, left[i]);
for (int k = 0; k < n; k++) {
int leftLen = (k > 0) ? left[k - 1] : 0;
int rightLen = (k + 1 < n) ? right[k + 1] : 0;
int cand = 1;
if (k > 0) cand = Math.max(cand, leftLen + 1);
if (k + 1 < n) cand = Math.max(cand, rightLen + 1);
if (k > 0 && k + 1 < n && serathion[k - 1] <= serathion[k + 1]) {
cand = Math.max(cand, leftLen + 1 + rightLen);
}
ans = Math.max(ans, cand);
}
return ans;
}
}