Optimal Path To Target
A Dynamic Programming Approach

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The DP problems belonging to this category, in its simplest form, looks like below or some kind of variations of it:
 Given a target find minimum (maximum) cost / path / sum to reach the target.
The solution for this kind of problems, in a very generalized form, would often look like below:
 Choose optimal (minimal or maximal, as the case may be) path among all possible paths that lead to the current state, and then add value for the current state.

routes[curr] = min(routes[curr  1], routes[curr  2], ... , routes[curr  k]) + cost[i]
where current target can be reached only from (curr  1), (curr  2), ... (curr  k). 
Overall the solution would look like this :
for (int curr = 1; curr <= target; curr++) { for (int k = 0; k < waysToReachCurrentTarget.size(); k++) { dp[i] = min(dp[curr], dp[waysToReachCurrentTarget[k]] + cost / path ) ; } } return dp[target];
The below problem and its solution beautifully demonstrate this approach:
Problem Statement:
Given a m x n grid filled with nonnegative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input:
[ [1,3,1], [1,5,1], [4,2,1] ]Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
Solution:
class Solution {
// 2D Dynamic Programming
public int minPathSum(int[][] grid) {
if (grid == null  grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = Integer.MAX_VALUE;
if (i == 0 && j == 0) {
dp[i][j] = grid[i][j];
} else {
if (i  1 >= 0) {
dp[i][j] = dp[i  1][j] + grid[i][j];
}
if (j  1 >= 0) {
dp[i][j] = Math.min(dp[i][j], dp[i][j  1] + grid[i][j]);
}
}
}
}
return dp[m  1][n  1];
}
}
Naive Solution:
int minCost(int cost[R][C], int m, int n)
{
if (n < 0  m < 0)
return INT_MAX;
else if (m == 0 && n == 0)
return cost[m][n];
else
return cost[m][n] + min( minCost(cost, m1, n1),
minCost(cost, m1, n),
minCost(cost, m, n1) );
}
Problem Solving:
The above content is written by:
Abhishek Dey
A Visionary Software Engineer With A Mission To Empower Every Person & Every Organization On The Planet To Achieve More
Microsoft  University of Florida
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