COWA19B-Editorial

COWA19B (Pongal Bunk)

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concept related: prefix array sum

Approach:-

Maintain two arrays (array1,array2).
-> first array (array1) to know the number of queries having the specific index in their range of L,R. For each query apply,

1)array1[L]=array1[L]+1
2) array1[R+1]=array1[R+1]-1

->second array (array2) to know the final value of element in the specific index of array for each query to apply,

1) array2[R+1]=array2[L]-(R-L+1)

After Performing All the queries operations,

->Run an iteration for prefix array sum for array1 i.e ,
array1[i]=array[i]+array1[i-1] (where 1<=i<=N)

->Run other iteration for array2 similar to prefix sum doing following operations
array2[i]=array2[i] + array2[i-1] + array1[i] (where 1<=i<=N)

Now, array2 contains the final values of elements present in array

Time-complexity: O(n+q+m)

reference-code for the problem is given below-


#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
using namespace std;
 
int main()
{
    ll ans[1000002],ans1[1000002];
    memset(ans1,0,sizeof(ans1));
    memset(ans,0,sizeof(ans));
    ll n,q,l,r;
    cin>>n>>q;
    while(q--)
    {
      cin>>l>>r;
      ans[l]+=1;ans[r+1]-=1;
      ans1[r+1]-=(r-l+1);
    }
    for(int i=1;i<=n;i++)
    ans[i]+=ans[i-1];
    for(int i=1;i<=n;i++)
    ans1[i]+=ans1[i-1]+ans[i];
    cin>>q;
    while(q--)
    {
      cin>>l;
      cout<<ans1[l]<<endl;
    }
} 

some-useful-links:-


@himkha_100
You have explained about how your solution works!
But could you also explain the intution behind how you came up with this solution!?

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