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# Weekly Contest 286 LeetCode Solution

## Problem 1 – Find the Difference of Two Arrays Leetcode Solution

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• -1000 <= nums1[i], nums2[i] <= 1000

### Find the Difference of Two Arrays Leetcode Solution in Python

def findDifference(self, nums1, nums2):
s1, s2 = set(nums1), set(nums2)
return [list(s1 - s2), list(s2 - s1)]

### Find the Difference of Two Arrays Leetcode Solution in C++

vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) {
vector<int> v1, v2;
set<int> s1(begin(nums1), end(nums1)), s2(begin(nums2), end(nums2));
set_difference(begin(s1), end(s1), begin(s2), end(s2), back_inserter(v1));
set_difference(begin(s2), end(s2), begin(s1), end(s1), back_inserter(v2));
return {v1, v2};
}

### Find the Difference of Two Arrays Leetcode Solution in Java

class Solution {
public List<List<Integer>> findDifference(int[] nums1, int[] nums2) {
Set<Integer> s1 = new HashSet<Integer>();
Set<Integer> s2 = new HashSet<Integer>();

for(int i : nums1){
}
for(int i : nums2){
}
List<List<Integer>> ret = new ArrayList();
for(int i : s1){
if(!s2.contains(i)){
}
}
for(int i : s2){
if(!s1.contains(i)){
}
}

return ret;
}
}

### Find the Difference of Two Arrays Leetcode Solution in Python3

class Solution:
def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
set1=set(nums1)
set2=set(nums2)
return [ list(set1-set2), list(set2-set1) ]

### Problem 2 – Minimum Deletions to Make Array Beautiful Leetcode Solution

You are given a 0-indexed integer array nums. The array nums is beautiful if:

• nums.length is even.
• nums[i] != nums[i + 1] for all i % 2 == 0.

Note that an empty array is considered beautiful.

You can delete any number of elements from nums. When you delete an element, all the elements to the right of the deleted element will be shifted one unit to the left to fill the gap created and all the elements to the left of the deleted element will remain unchanged.

Return the minimum number of elements to delete from nums to make it beautiful.

Example 1:

Input: nums = [1,1,2,3,5]
Output: 1
Explanation: You can delete either nums[0] or nums[1] to make nums = [1,2,3,5] which is beautiful. It can be proven you need at least 1 deletion to make nums beautiful.

Example 2:

Input: nums = [1,1,2,2,3,3]
Output: 2
Explanation: You can delete nums[0] and nums[5] to make nums = [1,2,2,3] which is beautiful. It can be proven you need at least 2 deletions to make nums beautiful.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 105

### Minimum Deletions to Make Array Beautiful Leetcode Solution in Python

def minDeletion(self, A):
res, pre = 0, -1
for a in A:
if a == pre:
res += 1
else:
pre = a if pre < 0 else -1
return res + (pre >= 0)

### Minimum Deletions to Make Array Beautiful Leetcode Solution in Java

public int minDeletion(int[] A) {
int res = 0, pre = -1;
for (int a : A) {
if (a == pre)
res++;
else
pre = pre < 0 ? a : -1;
}
return pre < 0 ? res : res + 1;
}

### Minimum Deletions to Make Array Beautiful Leetcode Solution in C++

int minDeletion(vector<int>& A) {
int res = 0, pre = -1;
for (int& a: A) {
if (a == pre)
res++;
else
pre = pre < 0 ? a : -1;
}
return res + (pre >= 0);
}

## Problem 3 – Find Palindrome With Fixed Length Leetcode Solution

Given an integer array queries and a positive integer intLength, return an array answer where answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1 if no such palindrome exists.

palindrome is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.

Example 1:

Input: queries = [1,2,3,4,5,90], intLength = 3
Output: [101,111,121,131,141,999]
Explanation:
The first few palindromes of length 3 are:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
The 90th palindrome of length 3 is 999.

Example 2:

Input: queries = [2,4,6], intLength = 4
Output: [1111,1331,1551]
Explanation:
The first six palindromes of length 4 are:
1001, 1111, 1221, 1331, 1441, and 1551.

Constraints:

• 1 <= queries.length <= 5 * 104
• 1 <= queries[i] <= 109
• 1 <= intLength <= 15

### Find Palindrome With Fixed Length Leetcode Solution in C++

int reverse(long long n, bool skip) {
long long res = 0;
for (n = skip ? n / 10 : n; n > 0; n /= 10)
res = res * 10 + n % 10;
return res;
}
vector<long long> kthPalindrome(vector<int>& queries, int sz) {
vector<long long> res;
long long start = pow(10, (sz + 1) / 2 - 1), end = pow(10, (sz + 1 ) / 2), mul = pow(10, sz / 2);
for (int q : queries)
res.push_back(start + q > end ? -1 :
(start + q - 1) * mul + reverse(start + q - 1, sz % 2));
return res;
}

### Find Palindrome With Fixed Length Leetcode Solution in Java

class Solution {
/*
First Palindrome of length 4 = "10"+"01"
First Palindrome of length 3 = "10"+"_1" (without first character 0)
First half can range from 10 (which is 10^1) to 99 (which is 10^2-1)
*/
public long[] kthPalindrome(int[] queries, int intLength) {
long[] result = new long[queries.length];
int i = 0;
for(int num: queries){
long half = (intLength+1)/2;//half for 3 =2, half for 4 = 2
long start = (long)Math.pow(10,half-1);
long end = (long)Math.pow(10,half)-1;
if(num<=(end-start+1)){ //check if query is within range
String firstHalfOfPalindrome = ((start)+ (num-1))+"";
String secondHalf = (new StringBuffer(firstHalfOfPalindrome)).reverse().toString();
result[i++] = Long.parseLong(firstHalfOfPalindrome+secondHalf.substring(intLength%2)); // since half is calculated as 1 extra for odd numbers, remove the first char for odd length in substring - 1001 vs 10+(0)1
}else{
result[i++]=-1;
}
}
return result;
}
}

### Find Palindrome With Fixed Length Leetcode Solution in Python

def kthPalindrome(self, queries, l):
base = 10 ** ((l - 1) / 2)
res = [q - 1 + base for q in queries]
for i,a in enumerate(res):
a = str(a) + str(a)[-1 - l % 2::-1]
res[i] = int(a) if len(a) == l else -1
return res

## Problem 4 – Maximum Value of K Coins From Piles Leetcode Solution

There are n piles of coins on a table. Each pile consists of a positive number of coins of assorted denominations.

In one move, you can choose any coin on top of any pile, remove it, and add it to your wallet.

Given a list piles, where piles[i] is a list of integers denoting the composition of the ith pile from top to bottom, and a positive integer k, return the maximum total value of coins you can have in your wallet if you choose exactly k coins optimally.

Example 1:

Input: piles = [[1,100,3],[7,8,9]], k = 2
Output: 101
Explanation:
The above diagram shows the different ways we can choose k coins.
The maximum total we can obtain is 101.

Example 2:

Input: piles = [[100],[100],[100],[100],[100],[100],[1,1,1,1,1,1,700]], k = 7
Output: 706
Explanation:
The maximum total can be obtained if we choose all coins from the last pile.

Constraints:

• n == piles.length
• 1 <= n <= 1000
• 1 <= piles[i][j] <= 105
• 1 <= k <= sum(piles[i].length) <= 2000

### Maximum Value of K Coins From Piles Leetcode Solution in Java

public int maxValueOfCoins(List<List<Integer>> piles, int k) {
Integer[][] memo = new Integer[piles.size() + 1][k + 1];

return dp(piles, memo, 0, k);
}
public int dp(List<List<Integer>> piles, Integer[][] memo, int i, int k) {
if (k == 0 || i == piles.size()) return 0;
if (memo[i][k] != null) return memo[i][k];

int res = dp(piles, memo, i + 1, k);
int cur = 0;

for (int j = 0; j < Math.min(piles.get(i).size(), k); ++j) {
cur += piles.get(i).get(j);
res = Math.max(res, cur + dp(piles, memo, i + 1, k - j - 1));
}
return memo[i][k] = res;
}

### Maximum Value of K Coins From Piles Leetcode Solution in C++

int maxValueOfCoins(vector<vector<int>>& A, int K) {
int n = A.size();
vector<vector<int>> memo(n + 1, vector<int>(K + 1, 0));
function<int(int, int)> dp = [&](int i, int k) {
if (memo[i][k] > 0) return memo[i][k];
if (i == n || k == 0) return 0;
int res = dp(i + 1, k), cur = 0;
for (int j = 0; j < A[i].size() && j < k; ++j) {
cur += A[i][j];
res = max(res, dp(i + 1, k - j - 1) + cur);
}
memo[i][k] = res;
return res;
};
return dp(0, K);
}

### Maximum Value of K Coins From Piles Leetcode Solution in Python

def maxValueOfCoins(self, A, K):

@functools.lru_cache(None)
def dp(i, k):
if k == 0 or i == len(A): return 0
res, cur = dp(i + 1, k), 0
for j in range(min(len(A[i]), k)):
cur += A[i][j]
res = max(res, cur + dp(i+1, k-j-1))
return res

return dp(0, K)
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