|
| 1 | +/* |
| 2 | +931. Minimum Falling Path Sum |
| 3 | +Medium |
| 4 | +
|
| 5 | +Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix. |
| 6 | +A falling path starts at any element in the first row and chooses the element in the next row that is |
| 7 | +either directly below or diagonally left/right. Specifically, |
| 8 | +the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1). |
| 9 | +
|
| 10 | + |
| 11 | +
|
| 12 | +Example 1: |
| 13 | +Input: matrix = [[2,1,3],[6,5,4],[7,8,9]] |
| 14 | +Output: 13 |
| 15 | +Explanation: There are two falling paths with a minimum sum as shown. |
| 16 | +
|
| 17 | +Example 2: |
| 18 | +Input: matrix = [[-19,57],[-40,-5]] |
| 19 | +Output: -59 |
| 20 | +Explanation: The falling path with a minimum sum is shown. |
| 21 | + |
| 22 | +
|
| 23 | +Constraints: |
| 24 | +
|
| 25 | +n == matrix.length == matrix[i].length |
| 26 | +1 <= n <= 100 |
| 27 | +-100 <= matrix[i][j] <= 100 |
| 28 | +*/ |
| 29 | +class Solution { |
| 30 | + static int[][] memo; |
| 31 | + public int minFallingPathSum(int[][] matrix) { |
| 32 | + |
| 33 | + int n = matrix.length; |
| 34 | + if (n == 1) return matrix[0][0]; |
| 35 | + //-100 <= matrix[i][j] <= 100 |
| 36 | + memo = new int[n+1][n+1]; |
| 37 | + Arrays.stream(memo).forEach(a->Arrays.fill(a,Integer.MAX_VALUE)); |
| 38 | + // int mini = Integer.MAX_VALUE; |
| 39 | + // for(int i=0;i<n;i++){ |
| 40 | + // mini = Math.min(mini , solveRec(0,i,n,matrix)); |
| 41 | + // } |
| 42 | + //return mini; |
| 43 | + |
| 44 | + return solveTab(matrix); |
| 45 | + |
| 46 | + //return solveOpt(matrix); |
| 47 | + } |
| 48 | + public int solveRec(int i,int j,int n,int[][] matrix){ |
| 49 | + //base case :: |
| 50 | + if(i==n){ |
| 51 | + return 0; |
| 52 | + } |
| 53 | + if(memo[i][j]!=Integer.MAX_VALUE){ |
| 54 | + return memo[i][j]; |
| 55 | + } |
| 56 | + |
| 57 | + int left =Integer.MAX_VALUE,right=Integer.MAX_VALUE; |
| 58 | + if(j>0){ |
| 59 | + left = solveRec(i+1,j-1,n,matrix); |
| 60 | + } |
| 61 | + int center = solveRec(i+1,j,n,matrix); |
| 62 | + if(j<n-1){ |
| 63 | + right = solveRec(i+1,j+1,n,matrix); |
| 64 | + } |
| 65 | + return memo[i][j] = matrix[i][j] + Math.min(left,Math.min(center,right)); |
| 66 | + } |
| 67 | + public int solveTab(int[][] matrix){ |
| 68 | + int n = matrix.length; |
| 69 | + int[][] tab = new int[n][n]; |
| 70 | + |
| 71 | + for(int i=0;i<n;i++){ |
| 72 | + tab[0][i] = matrix[0][i]; |
| 73 | + } |
| 74 | + |
| 75 | + for(int i=1;i<n;i++){ |
| 76 | + for(int j=0;j<n;j++){ |
| 77 | + int left =Integer.MAX_VALUE,right=Integer.MAX_VALUE; |
| 78 | + if(j>0){ |
| 79 | + left = tab[i-1][j-1]; |
| 80 | + } |
| 81 | + int center = tab[i-1][j]; |
| 82 | + if(j<n-1){ |
| 83 | + right = tab[i-1][j+1]; |
| 84 | + } |
| 85 | + tab[i][j] = matrix[i][j] + Math.min(left,Math.min(center,right)); |
| 86 | + } |
| 87 | + } |
| 88 | + |
| 89 | + int maxi = Integer.MAX_VALUE; |
| 90 | + for (int j = 0; j < n; j++) { |
| 91 | + maxi = Math.min(maxi, tab[n - 1][j]); |
| 92 | + } |
| 93 | + |
| 94 | + return maxi; |
| 95 | + } |
| 96 | + public int solveOpt(int[][] matrix){ |
| 97 | + int n = matrix.length; |
| 98 | + int[] prev = new int[n]; |
| 99 | + |
| 100 | + for(int i=0;i<n;i++){ |
| 101 | + prev[i] = matrix[0][i]; |
| 102 | + } |
| 103 | + |
| 104 | + for(int i=1;i<n;i++){ |
| 105 | + int[] temp = new int[n]; |
| 106 | + for(int j=0;j<n;j++){ |
| 107 | + int left =Integer.MAX_VALUE,right=Integer.MAX_VALUE; |
| 108 | + if(j>0){ |
| 109 | + left = prev[j-1]; |
| 110 | + } |
| 111 | + int center = prev[j]; |
| 112 | + if(j<n-1){ |
| 113 | + right = prev[j+1]; |
| 114 | + } |
| 115 | + temp[j] = matrix[i][j] + Math.min(left,Math.min(center,right)); |
| 116 | + } |
| 117 | + prev = temp; |
| 118 | + } |
| 119 | + |
| 120 | + int maxi = Integer.MAX_VALUE; |
| 121 | + for (int j = 0; j < n; j++) { |
| 122 | + maxi = Math.min(maxi, prev[j]); |
| 123 | + } |
| 124 | + |
| 125 | + return maxi; |
| 126 | + } |
| 127 | +} |
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