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MinimumFallingPathSum.cpp
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MinimumFallingPathSum.cpp
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// Source : https://leetcode.com/problems/minimum-falling-path-sum/
// Author : Hao Chen
// Date : 2019-01-30
/*****************************************************************************************************
*
* Given a square array of integers A, we want the minimum sum of a falling path through A.
*
* A falling path starts at any element in the first row, and chooses one element from each row. The
* next row's choice must be in a column that is different from the previous row's column by at most
* one.
*
* Example 1:
*
* Input: [[1,2,3],[4,5,6],[7,8,9]]
* Output: 12
* Explanation:
* The possible falling paths are:
*
* [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
* [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
* [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
*
* The falling path with the smallest sum is [1,4,7], so the answer is 12.
*
* Note:
*
* 1 <= A.length == A[0].length <= 100
* -100 <= A[i][j] <= 100
******************************************************************************************************/
class Solution {
private:
int min(int x, int y) {
return x < y ? x: y;
}
int min( int x, int y, int z) {
return min(min(x, y),z);
}
public:
int minFallingPathSum(vector<vector<int>>& A) {
int m = INT_MAX;
for (int i=0; i<A.size(); i++) {
for (int j=0; j<A[i].size(); j++){
//find the minimal item in previous row, and add it into the current item
if (i > 0) {
if (j == 0 ){
A[i][j] += min( A[i-1][j], A[i-1][j+1]);
} else if ( j + 1 == A[i].size()) {
A[i][j] += min( A[i-1][j], A[i-1][j-1]);
}else {
A[i][j] += min( A[i-1][j], A[i-1][j-1], A[i-1][j+1]);
}
}
if ( i + 1 == A.size() ) {
m = min(m, A[i][j]);
}
}
}
return m;
}
};