**Maximum Dropping Path Sum With Column Distance Cost - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers matrix. You want to pick a number from each row. For each 0 ≤ r < n - 1 the cost for picking matrix[r][j] and matrix[r + 1][k] is abs(k - j). Return the maximum sum possible of the numbers chosen, minus costs. Constraints 1 ≤ n * m ≤ 200,000 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [3, 2, 1, 6], [4, 1, 2, 0], [1, 5, 2, -2] ] Output 11 Explanation Return 11 by picking 3, 4 and 5. Example 2 Input matrix = [ [7, 6, 5, 6], [6, 4, 5, 8] ] Output 14 Explanation We can take the last 6 in the first row and 8 in the last row. Example 3 Input matrix = [ [2, 1, 3] ] Output 3 Explanation We can take 3.

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
int n = matrix.size();
int m = matrix[0].size();
vector<vector<int>> dp(n, vector<int>(m, INT_MIN));
vector<int> left(m, INT_MIN), right(m, INT_MIN);
for (int i = 0; i < m; i++) {
dp[0][i] = matrix[0][i];
}
for (int i = 0; i < m; i++) {
if (i > 0) {
left[i] = max(left[i - 1], dp[0][i] + i);
} else {
left[0] = dp[0][0];
}
int ri = m - 1 - i;
if (ri < m - 1) {
right[ri] = max(right[ri + 1], dp[0][ri] - ri);
} else {
right[ri] = dp[0][ri] - ri;
}
}
for (int i = 1; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j] = max({dp[i][j], left[j] + matrix[i][j] - j, right[j] + matrix[i][j] + j});
}
for (int j = 0; j < m; j++) {
left[j] = INT_MIN;
if (j > 0) {
left[j] = max(left[j - 1], dp[i][j] + j);
} else {
left[0] = dp[i][0];
}
int ri = m - 1 - j;
right[ri] = INT_MIN;
if (ri < m - 1) {
right[ri] = max(right[ri + 1], dp[i][ri] - ri);
} else {
right[ri] = dp[i][ri] - ri;
}
}
}
int ans = *max_element(dp[n - 1].begin(), dp[n - 1].end());
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int N = matrix.length;
int M = matrix[0].length;
int NEGINF = Integer.MIN_VALUE;
int[][] dp = new int[N][M];
for (int j = 0; j < M; j++) {
dp[0][j] = matrix[0][j];
}
for (int i = 1; i < N; i++) {
int[] right = new int[M + 1];
right[M] = NEGINF;
for (int j = M - 1; j >= 0; j--) right[j] = Math.max(right[j + 1], dp[i - 1][j] - j);
int left = NEGINF;
for (int j = 0; j < M; j++) {
left = Math.max(left, dp[i - 1][j] + j);
dp[i][j] = matrix[i][j] + Math.max(left - j, right[j] + j);
}
}
int ans = NEGINF;
for (int j = 0; j < M; j++) {
ans = Math.max(ans, dp[N - 1][j]);
}
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, matrix):
dp = matrix[0]
for i in range(1, len(matrix)):
ndp = [float(-inf)] * len(matrix[i])
lhs = float(-inf)
for j in range(len(matrix[i])):
lhs = max(lhs, dp[j])
ndp[j] = max(ndp[j], lhs + matrix[i][j])
lhs -= 1
rhs = float(-inf)
for j in range(len(matrix[i]) - 1, -1, -1):
rhs = max(rhs, dp[j])
ndp[j] = max(ndp[j], rhs + matrix[i][j])
rhs -= 1
dp = ndp
return max(dp)
```

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