# WRDVLS - Editorial

Setter: Nishan Singh
Tester: Aryan Choudhary
Editorialist: Ajit Sharma Kasturi

EASY

None

### PROBLEM:

Given and array A of N integers, the weirdness of the subarray A_L, A_{L+1}, \dots , A_R denoted by w(L, R) is defined as the number of indices L \leq i \leq R for which A_i equals the number of occurrences of A_i in that subarray. We need to find the value of \sum_{L=1}^{N} \sum_{R=L}^{N} w(L, R) .

### EXPLANATION:

• First let us create an array vals_x for every unique value x present in A. vals_x denotes the array of integers i_1, i_2, i_3, \dots i_{cnt_x} for which A_{i_1} = A_{i_2} = \dots = A_{i_{cnt_x}} = x. Here, cnt_x denotes the total number of values in A equal to x.

• Suppose we fix some x. Let us consider a technique of sliding window with window of size x and keep sliding the window on the array vals_x.

• Suppose our window is like i_j, i_{j+1}, \dots, i_{j+x-1}. Now let us solve this problem which is, how much does this window contribute to the answer?

• To solve this, we need to find the number of subarrays which contain this window and have no other index k apart from those in window with A_k = x.

• Clearly, the starting point of such subarray can be in the range
[i_{j-1}+1, i_j] and the ending point of such subarray can be in the range [i_{j+x-1}, i_{j+x} -1]. ( If i_{j-1} doesn’t exist, we can replace it with 0 and if i_{j+x} doesn’t exist, we can replace it with N+1 ).

• Therefore, the total number of such subarrays for the current window are
(i_j - i_{j-1}) \cdot (i_{j+x} - i_{j+x-1}).

• Each subarray have x values equal to x, so each subarray contributes x to the answer. So, the total contribution will be x \cdot (i_j - i_{j-1}) \cdot (i_{j+x} - i_{j+x-1}).

• We can similarly do this for every window of every unique value x in the array A and add its contribution to the answer.

### TIME COMPLEXITY:

O(N) for each testcase.

### SOLUTION:

Editorialist's solution
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

ll solve(vector<int>& vals, int num, int original_array_size) {
int n = vals.size();
ll ans = 0;
for (int i=0; i<n; i++) {
if (i+num-1 >= n) {
break;
}

int l = vals[i], r = vals[i+num-1];
int prev_l = -1, next_r = original_array_size;
if(i-1 >= 0) prev_l = vals[i-1];
if(i+num < n) next_r = vals[i+num];

ans += 1LL * (l - prev_l) * (next_r - r) * num;
}

return ans;
}

int main()
{
int tests;
cin >> tests;
while (tests--) {
int n;
cin >> n;
vector<int> a(n);
unordered_map<int, vector<int>> vals;

for(int i=0; i<n; i++) {
cin >> a[i];
vals[a[i]].push_back(i);
}

ll ans = 0;
for(auto p: vals) {
ans += solve(p.second, p.first, n);
}

cout << ans << endl;
}
return 0;
}

Setter's solution
#include<bits/stdc++.h>
using namespace std;
#define int long long

int calc(vector<int> &v, int num, int n){
if(v.size()<num)return 0;

int res = 0;
for(int i = 0; i+num-1 < v.size(); i++){
int pre, nex;
if(i>0)pre = v[i] - v[i-1];
else pre = v[i]+1;

if(i+num<v.size())nex = v[i+num] - v[i+num-1];
else nex = n-v[i+num-1];
res += num*pre*nex;
}
return res;
}

vector<int> v[1000001];

void solve(){
int n;
cin >> n;
int a[n];

for(int i = 0; i<n; i++){
cin >> a[i];
if(a[i]<=n){
v[a[i]].push_back(i);
}
}

int ans = 0;
for(int i = 1; i<=n; i++){
ans += calc(v[i], i, n);
}
for(int i = 0; i<n; i++){
v[i].clear();
}

cout << ans << "\n";

}

signed main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);

#ifndef ONLINE_JUDGE
freopen("./testgen/Tests/inputs/input1.txt", "r", stdin);
freopen("./testgen/Tests/outputs/output1.txt", "w", stdout);
#endif

int test=0;
cin >> test;

while(test--){solve();}

return 0;
}

Tester's solution
def main():
for _ in range(int(input())):
n=int(input())
a=[int(x) for x in input().split()]
b=[[] for _ in range(n+1)]
for i,x in enumerate(a):
if x <=n:
b[x].append(i)
def solve(a,x):
m=len(a)
if m==0:
return 0
ans=0
for i in range(m-x+1):
L,R=i,i+x-1
l=a[L]-(a[L-1] if L else -1)
r=(a[R+1] if R+1<m else n)-a[R]
ans+=l*r
# print(ans,x)
return ans
print(sum(solve(v,i)*i for i,v in enumerate(b)))

main()



Please comment below if you have any questions, alternate solutions, or suggestions.

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