Minimum Number of Operations to Satisfy Conditions
<p>You are given a 2D matrix <code>grid</code> of size <code>m x n</code>. In one <strong>operation</strong>, you can change the value of <strong>any</strong> cell to <strong>any</strong> non-negative number. You need to perform some <strong>operations</strong> such that each cell <code>grid[i][j]</code> is:</p>

<ul>
	<li>Equal to the cell below it, i.e. <code>grid[i][j] == grid[i + 1][j]</code> (if it exists).</li>
	<li>Different from the cell to its right, i.e. <code>grid[i][j] != grid[i][j + 1]</code> (if it exists).</li>
</ul>

<p>Return the <strong>minimum</strong> number of operations needed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1,0,2],[1,0,2]]</span></p>

<p><strong>Output:</strong> 0</p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3122.Minimum%20Number%20of%20Operations%20to%20Satisfy%20Conditions/images/examplechanged.png" style="width: 254px; height: 186px;padding: 10px; background: #fff; border-radius: .5rem;" /></strong></p>

<p>All the cells in the matrix already satisfy the properties.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1,1,1],[0,0,0]]</span></p>

<p><strong>Output:</strong> 3</p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3122.Minimum%20Number%20of%20Operations%20to%20Satisfy%20Conditions/images/example21.png" style="width: 254px; height: 186px;padding: 10px; background: #fff; border-radius: .5rem;" /></strong></p>

<p>The matrix becomes <code>[[1,0,1],[1,0,1]]</code> which satisfies the properties, by doing these 3 operations:</p>

<ul>
	<li>Change <code>grid[1][0]</code> to 1.</li>
	<li>Change <code>grid[0][1]</code> to 0.</li>
	<li>Change <code>grid[1][2]</code> to 1.</li>
</ul>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1],[2],[3]]</span></p>

<p><strong>Output:</strong> 2</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3122.Minimum%20Number%20of%20Operations%20to%20Satisfy%20Conditions/images/changed.png" style="width: 86px; height: 277px;padding: 10px; background: #fff; border-radius: .5rem;" /></p>

<p>There is a single column. We can change the value to 1 in each cell using 2 operations.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, m &lt;= 1000</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 9</code></li>
</ul>
Array; Dynamic Programming; Matrix
class Solution {
public:
    int minimumOperations(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int f[n][10];
        memset(f, 0x3f, sizeof(f));
        for (int i = 0; i < n; ++i) {
            int cnt[10]{};
            for (int j = 0; j < m; ++j) {
                ++cnt[grid[j][i]];
            }
            if (i == 0) {
                for (int j = 0; j < 10; ++j) {
                    f[i][j] = m - cnt[j];
                }
            } else {
                for (int j = 0; j < 10; ++j) {
                    for (int k = 0; k < 10; ++k) {
                        if (k != j) {
                            f[i][j] = min(f[i][j], f[i - 1][k] + m - cnt[j]);
                        }
                    }
                }
            }
        }
        return *min_element(f[n - 1], f[n - 1] + 10);
    }
};