Minimum Operations to Make Columns Strictly Increasing

• The core idea is that operations on one column are independent of others. We can calculate the minimum operations for each column and sum them up.
• For a single column, we can use dynamic programming. Let dp[i][v] be the minimum cost to make the prefix of the column of length i+1 (i.e., col[0...i]) strictly increasing, with the final value of col[i] being v.
• The state transition is: dp[i][v] = (v - col[i]) + min_{u < v} {dp[i-1][u]}. This means the cost is the operations to change col[i] to v, plus the minimum cost for the previous i-1 elements to be strictly increasing, ending in a value u that is less than v.
• The base case for i=0 is dp[0][v] = v - col[0] for all v >= col[0].
• To make the computation efficient, we can optimize finding min_{u < v} {dp[i-1][u]}. For each row i, we can pre-calculate a running minimum of the dp[i-1] row. This reduces the complexity of calculating each dp[i][v] from O(V_max) to O(1).
• The final answer for a column is the minimum value in the last row of the DP table, min(dp[m-1]).
• We can optimize space by only keeping track of the previous and current rows of the DP table.

The problem can be broken down by columns, as the modifications in one column do not affect any other. The total minimum operations is the sum of minimum operations required for each individual column.

For a single column, we can formulate a dynamic programming solution. Let dp[i][v] be the minimum number of operations to make the first i+1 elements (from row 0 to i) of the column strictly increasing, with the element at row i having the final value v.

The state transition is derived as follows: to have the element at row i become v, we need v - grid[i][j] operations (assuming v >= grid[i][j]). The element at row i-1 must have a final value u that is strictly less than v. To minimize the total cost, we should choose the u that resulted in the minimum operations for the prefix up to i-1. Thus, the recurrence relation is: dp[i][v] = (v - grid[i][j]) + min_{u < v} {dp[i-1][u]}.

The base case is for the first row (i=0): dp[0][v] = v - grid[0][j] for v >= grid[0][j].

The range for v needs to be determined. A safe upper bound is the maximum possible initial value in the grid plus the number of rows m, as in the worst case, we might need to increment by 1 for each subsequent row. Let's call this V_max.

A naive implementation of the transition would be slow. We can optimize the calculation of min_{u < v} {dp[i-1][u]}. For each row i, we can compute a running minimum of the dp[i-1] values. This allows us to find the minimum cost for the previous state in O(1) time.

We can also optimize space from O(m * V_max) to O(V_max) by only storing the DP results for the previous and current rows.

1import java.util.Arrays;2 3class Solution {4    public int minOperations(int[][] grid) {5        int m = grid.length;6        int n = grid[0].length;7        long totalOperations = 0;8 9        int maxInitialVal = 0;10        for (int i = 0; i < m; i++) {11            for (int j = 0; j < n; j++) {12                maxInitialVal = Math.max(maxInitialVal, grid[i][j]);13            }14        }15        16        int V_MAX = maxInitialVal + m;17 18        for (int j = 0; j < n; j++) {19            long[] prevDp = new long[V_MAX];20            Arrays.fill(prevDp, Long.MAX_VALUE);21 22            // Base case: i = 023            for (int v = grid[0][j]; v < V_MAX; v++) {24                prevDp[v] = v - grid[0][j];25            }26 27            // Iterate through rows i = 1 to m-128            for (int i = 1; i < m; i++) {29                long[] currentDp = new long[V_MAX];30                Arrays.fill(currentDp, Long.MAX_VALUE);31                32                long minPrevCost = Long.MAX_VALUE;33                for (int v = 1; v < V_MAX; v++) {34                    minPrevCost = Math.min(minPrevCost, prevDp[v - 1]);35                    if (v >= grid[i][j] && minPrevCost != Long.MAX_VALUE) {36                        currentDp[v] = (long)(v - grid[i][j]) + minPrevCost;37                    }38                }39                prevDp = currentDp;40            }41 42            long minColOps = Long.MAX_VALUE;43            for (long ops : prevDp) {44                minColOps = Math.min(minColOps, ops);45            }46            if (minColOps != Long.MAX_VALUE) {47                totalOperations += minColOps;48            }49        }50 51        return (int)totalOperations;52    }53}