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I would categorize them as leetcode medium

• I would appretiate if someone post the coding soltion for the 1st question, as my only 4 testcases passed during round.

by Expert (111,330 points)
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Q1.

``````public List<Integer> GOODARRAY(int N, List<List<Integer>> queries){
ArrayList<Integer> goodArray = new ArrayList<>();
oneBitCount(goodArray, N);

long goodArrayProduct = 1;

for(List<Integer> query : queries)
{
int l = query.get(0);
int r = query.get(1);
int m = query.get(2);
for(int i=l; i<=r; i++) goodArrayProduct*=goodArray.get(i-1);
}
}
public void oneBitCount(ArrayList<Integer> goodArray, int n)
{
int count = 0;
while(n > 0)
{
if((n & 1) == 1) goodArray.add((int)Math.pow(2,count));
count++;
n = n>>1;
}
}
``````

Q1.
Step 1 :- find the position of set bits in binary representation of number N and store it in a array lets call it nums;
Step 2 :- prepare prefix sum of array nums
Step 3 :- now traverse every query in queries and store your answer as ------- pow(2, prefixNums[right] - prefixNums[left - 1]) % modulo ;

For Q1.A similar idea has been given in the comments. I too came up with that idea and have used binary exponentiation to speed things.
For Q2. you just need to sort the array a and then do binary search / upper bound to find the index of the last match that was equal to or less than the goals scored in b's match.

``````[#include](https://leetcode.com/problems/minimum-interval-to-include-each-query) <bits/stdc++.h>
using namespace std;

int binpow(int a, int n, int mod){
if(a == 1)return 1;
if(n == 0)return 1;
if(n == 1)return a%mod;
if(n&1)return binpow(a, n/2, mod)%mod *(binpow(a, n/2, mod)%mod * a%mod)%mod;
return binpow(a, n/2, mod)%mod * binpow(a, n/2, mod)%mod;
}

void solve(int n, vector<vector<int>>queries){
vector<int>arr;
int _n = n;
int cnt = 0;
while(_n){
if(_n&1){
arr.push_back(cnt);
}
++cnt;
_n = _n/2;
}

vector<long long>prefix;
prefix.push_back(0);

for(int i = 1; i<=(int)arr.size();++i){
prefix.push_back(prefix[i-1] + arr[i-1]);
}
for(auto&it:queries){
int l = it[0], r = it[1], mod = it[2];
int power = prefix[r] - prefix[l-1];
long long ans = binpow(2, power, mod);
cout<<ans<<"\n";
}
return;
}

int main() {
int n = 26;
vector<vector<int>>queries= {{1, 2, 1009}, {3, 3, 5}};
solve(n, queries);
return 0;
}
``````

`FOR Q2`

``````[#include](https://leetcode.com/problems/minimum-interval-to-include-each-query) <bits/stdc++.h>
using namespace std;

vector<int> solve(vector<int>&a, vector<int>&b){
sort(a.begin(), a.end());
// number of matches with less than or equal to.
vector<int>ans;
int n = (int)a.size();
for(auto&it:b){
int idx = upper_bound(begin(a), end(a), it) - begin(a);
if(idx == n){
ans.push_back(n);
}else{
ans.push_back(idx);
}
}
return ans;
}
int main() {
vector<int>a = {1, 2, 4};
vector<int>b = {2, 4};
vector<int>ans = solve(a, b);
for(auto&it:ans)cout<<it<<" ";
cout<<"\n";
return 0;
}``````
by Expert (111,330 points)