Question:

You have a $$$N * M$$$ matrix. The color of cell in $$$i^{th}$$$ row and $$$j^{th}$$$ column is $$$a_{ij}$$$. Find the number of rectangles that can be formed from this matrix by deleting some rows and columns, that have all cells of same color.

Idea:

Let $$$i_l, j_l$$$ be the coordinates of the top left corner and $$$i_r, j_r$$$ be the coordinates of the bottom right corner. We can see that $$$(i_l, j_l, i_r, j_r)$$$ uniquely define a rectangle.

Let's call a rectangle valid if it has all cells of same color.

Now let $$$dp[i][j]$$$ be the number of valid rectangles with their bottom right corner at $$$(i, j)$$$.

Let $$$h[i][j]$$$ be the "height" of columns at $$$(i, j)$$$. That is

$$$h[i][j]$$$ = max value of $$$len$$$ such that $$$a[i][j] = a[i][t]$$$ , for all $$$t = i - len + 1, i - len + 2, ..., i$$$.

How do we computer $$$dp[i][j]$$$ efficiently ?

Consider how $$$dp[i][j]$$$ depends on $$$dp[i-1][j], dp[i - 2][j], ...$$$.

For some $$$k, k < j$$$, if $$$a[i][k] != a[i][j]$$$, then $$$dp[i][k]$$$ will make no contribution to $$$dp[i][j]$$$. So let's assume that $$$a[i][t] = a[i][j]$$$ for all $$$t = k, k + 1, ..., j$$$.

Now what is the maximum height of rectangle with bottom right corner at $$$(i, j)$$$ and top left corner in the column $$$k$$$ ?

It is $$$height_k = min_{k \le t \le i}(h[i][t])$$$. Then then the number of such rectangles is $$$height_k$$$, since we can have rectangles of height $$$1, 2, 3, ..., height_k$$$.

Also consider that if $$$h[i][k] \le height_k$$$, then for any rectangle with bottom right corner at $$$(i, k)$$$ can be extended so that it has bottom right corner at $$$(i, j)$$$.

Let $$$k$$$ be the first index such that $$$a[i][k] = a[i][j]$$$ and $$$h[i][k] < h[i][j]$$$.

Then every rectangle with bottom right corner at $$$(i, k)$$$ can be extended to $$$(i, j)$$$. Also the height of any rectangle with top-left corner after $$$(i, k)$$$ is at most $$$h[i][j]$$$, since by definition for any $$$t, k < t \le i$$$ we have $$$h[i][t] \ge h[i][j]$$$.

Thus the number of rectangles after k are $$$h[i][j] * (j - k)$$$.

Thus we have, $$$dp[i][j] = dp[i][k] + h[i][j] * (j - k)$$$.

Now the question is how do we efficiently find $$$k$$$ ?

We can do this by maintaining an array of indices $$$col$$$ such that $$$h[i][col] \lt min_{col \le t \le j}(h[i][t])$$$.

Time Complexity : $$$O(N * M)$$$

vector<vector<ll>> get_number_of_rectangles(vector<vector<int>> &a, int n, int m){
    vector<vector<ll>> dp(n, vector<ll>(m));
    vector<vector<ll>> h(n, vector<ll>(m, 1));

    for(int i = 1; i < n; ++i){
        for(int j = 0; j < m; ++j){
            if(a[i - 1][j] == a[i][j]){
                h[i][j] += h[i - 1][j];
            }
        }
    }

    for(int i = 0; i < n; ++i){
        vector<int> indices;
        for(int j = 0; j < m; ++j){
            int k = -1;
            while(!indices.empty()){
                int col = indices.back();
                if((a[i][col] != a[i][j]) || (h[i][col] < h[i][j])){
                    break;
                }
                indices.pop_back();
            }
            if(!indices.empty()){
                k = indices.back();
            }
            dp[i][j] += h[i][j] * (j - k);
            if(k != -1){
                if(a[i][k] == a[i][j]){
                    dp[i][j] += dp[i][k];
                }
            }
            indices.push_back(j);
        }
    }

    return dp;
}
 
void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> a(n, vector<int>(m));
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            cin >> a[i][j];
        }
    }
    vector<vector<ll>> dp = get_number_of_rectangles(a, n, m);
    ll ans = 0;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < m; ++j){
            ans += dp[i][j];
        }
    }
    cout << ans << '\n';
}