## A - Remove One Character Editorial by evima

In this editorial, let \(S(k)\) denote the \(k\)-th character of \(S\).

### ◆ Rephrasing the condition \(S_i = S_j\)

First, let us figure out when \(S_i = S_j\) holds for \(1\leq i < j \leq N\).

The equation \(S_i = S_j\) holds if and only if \(S_i(k) = S_j(k)\) holds for every \(k\). Let us examine where the \(k\)-th characters of \(S_i, S_j\) are located in \(S\).

- If \(1\leq k < i\): we have \(S_i(k) = S(k)\), \(S_j(k) = S(k)\).
- If \(i\leq k < j\): we have \(S_i(k) = S(k+1)\), \(S_j(k) = S(k)\).
- If \(j\leq k < N\): we have \(S_i(k) = S(k+1)\), \(S_j(k) = S(k+1)\).

Thus, it can be seen that \(S_i=S_j \iff S(i) = S(i+1) = \cdots = S(j)\).

### ◆ Counting

Let us count the pairs \((i,j)\) for each \(j\) separately. We want to compute the following for each \(j\):

find the number \(n_j\) of indices \(i<j\) such that \(S(i) = \cdots = S(j)\).

First, if \(S(j-1)\neq S(j)\), we have \(n_j = 0\). If \(S(j-1)=S(j)\), the condition can be rephrased into “\(i=j-1\), or \(i<j-1\) and \(S(i) = \cdots = S(j-1)\)”, so we have \(n_j = n_{j-1} + 1\).

Thus, by computing \(n_j\) in the order \(j=1, 2, \ldots\), each \(n_j\) can be computed in \(O(1)\) time. Therefore, the problem can be solved in \(O(N)\) time.

### ◆Sample Implementation

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