WhiteHat Coding
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class Solution{
public:
#define ll long long
long long noOfPairs(vector<string> &box){
vector<ll>A;
for(auto it : box){
vector<int>D(26 ,0);
for(auto x : it){
D[x-'a']++;
}
long long val = 0;
for(int i=0;i<26;i++){
if(D[i]%2) val += (1ll<<i);
}
A.push_back(val);
}
//we want to find number of pair such that
//xor of these 2 has atmost one bit set
unordered_map<ll ,ll>mp;
for(auto it : A){
mp[it]++;
}
ll ans =0 ;
for(auto it : A){
mp[it]--;
if(mp[it]==0) mp.erase(mp.find(it));
for(int i=0;i<26;i++){
//we want resultant be (1ll<<i)
ll want = (1ll<<i);
ll want_mask = (want^it);
if(mp.count(want_mask)) ans += mp[want_mask];
}
//also we want no body also
if(mp.count(it)) ans += mp[it];
}
return ans;
}
};
class Solution{
public int maxGoodLength(int arr[][]){
// Code Here.
int N = arr.length;
int M = arr[0].length;
int low = 0;
int high = Math.min(N, M);
int maxGoodLength = 0;
while (low <= high) {
int mid = (low + high + 1) / 2;
boolean found = false;
for (int i = 0; i <= N - mid; i++) {
for (int j = 0; j <= M - mid; j++) {
boolean valid = true;
for (int x = i; x < i + mid; x++) {
for (int y = j; y < j + mid; y++) {
if (arr[x][y] < mid) {
valid = false;
break;
}
}
if (!valid) {
break;
}
}
if (valid) {
found = true;
break;
}
}
if (found) {
break;
}
}
if (found) {
maxGoodLength = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
return maxGoodLength;
}
}
class Solution{
public:
int minimumMagic(int n, int m, vector<int> &p, vector<int> &mp)
{
priority_queue<int> pq;
int sum = 0;
for(auto x : mp) sum += x;
if(sum > m) return -1;
sum = 0;
for(auto x : p) sum += x;
for(int i = 0; i < n; i++) {
pq.push(p[i] - mp[i]);
}
int ans = 0;
while(pq.size()){
if(sum <= m) return ans;
sum -= pq.top();
pq.pop();
ans++;
}
return ans;
}
};
Guys plagiarism will be checked so please change method order and variable name
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