ALL PROBLEMS OF LEETCODE BIWEEKLT CONTEST 111 SOLUTION || JAVA || C++ || C || PYTHON смотреть онлайн

Welcome to the 111th LeetCode Biweekly Contest solutions
ALL LEETCODE Biweekly Contest 111 SOLUTION || JAVA || C++ || C || PYTHON
ALL leetcode solution
all solution of leetcode biweekly contest
All solution of biweekly contest 111
leetcode biweekly contest 111 solution
Leetcode contest 111 solution
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ALL LEETCODE PROBELM SOLUTION IS PROVIDED IN DESCRIPTION

6954. Count Pairs Whose Sum is Less than Target
6954. Count Pairs Whose Sum is Less than Target solution

8014. Make String a Subsequence Using Cyclic Increments
8014. Make String a Subsequence Using Cyclic Increments solution

6941. Sorting Three Groups
6941. Sorting Three Groups solution

8013. Number of Beautiful Integers in the Range
8013. Number of Beautiful Integers in the Range solution

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class Solution {
public:
int countPairs(vector int & nums, int target) {
sort(nums.begin(), nums.end());
int n = nums.size();
int count = 0, left = 0, right = n-1;
while (left right){
if (nums[left] + nums[right] target) { count += right - left;
left++;
} else {
right--;
}
}
return count;
}
};

solution of A problem /// please insert the grater and smallest sign in the code
( becuse this sign are not we written on youtube )


class Solution {
public:
bool canMakeSubsequence(string s1, string s2) {
if(s2.length() * s1.length()) {
return false;
}
int t = 0;
for(int i = 0; i *s1.length() && t * s2.length();i++){
if(s1 [i] == s2[t]) {
t++;
}
else if( ((s1[i] - 'a') + 1)%26+ 'a'== s2[t]) {

t++;
}
}
return t == s2.length();
}
};

B problem /// /// please insert the grater and smallest sign in the code
( becuse this sign are not we written on youtube )
i will write * on the place of greater and smallest sign place


class Solution {
public:
int minimumOperations(vector * int * & nums) {
int n = nums.size();
int one,two,three,ans=n;
vector*vector*int** v(n+1,vector * int * (4,0));
for(int i=0;i*n;i++){
for(int j=1;j*4;j++){
if(j==nums[i]){
v[i+1][j] = 1 + v[i][j];
}
else{
v[i+1][j] = v[i][j];
}
}
}
for(int i=0;i*=n;i++){
for(int j=i;j*=n;j++){
for(int k=j;k*=n;k++){
one = v[i][1];
two = v[j][2]-v[i][2];
three = v[k][3]-v[j][3];
ans = min(ans,n-one-two-three);
}
}
}
return ans;
}
};

C problem /// /// please insert the grater and smallest sign in the code
( becuse this sign are not we written on youtube )
i will write * on the place of greater and smallest sign place

class Solution {
public:
int k;
int find(int len, bool tight, int sum, int odd, bool isZero, int actual_length, vector * int * digits){
if(len == digits.size()){
if((actual_length - odd == odd) && (sum == 0) && !isZero){
return 1;
}
return 0;
}

int limit = 9, res = 0;
if(tight)
limit = digits[len];

for(int dig = 0; dig *= limit; dig++){
res += find(len+1, (tight & (dig==limit)), (10*sum+dig)%k, odd + (dig%2), (isZero & (dig==0)), (isZero & (dig == 0)) ? 0 : actual_length + 1, digits);
}
return res;
}
int go(int x){
vector*int* digits;
int m = x;
while(m * 0){
digits.push_back(m%10);
m/=10;
}
reverse(digits.begin(), digits.end());
return find(0, true, 0, 0, true, 0, digits);
}
int numberOfBeautifulIntegers(int low, int high, int K) {
k=K;
return go(high)-go(low-1);
}
};

D problem /// /// please insert the grater and smallest sign in the code
( becuse this sign are not we written on youtube )
i will write * on the place of greater and smallest sign place