Thursday, June 23, 2016

LeetCode 363 - Max Sum of Rectangle No Larger Than K


https://leetcode.com/problems/max-sum-of-sub-matrix-no-larger-than-k/
Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.
Example:
Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2
The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).
Note:
  1. The rectangle inside the matrix must have an area > 0.
  2. What if the number of rows is much larger than the number of columns?
http://bookshadow.com/weblog/2016/06/22/leetcode-max-sum-of-sub-matrix-no-larger-than-k/
题目可以通过降维转化为求解子问题:和不大于k的最大子数组,解法参考:https://www.quora.com/Given-an-array-of-integers-A-and-an-integer-k-find-a-subarray-that-contains-the-largest-sum-subject-to-a-constraint-that-the-sum-is-less-than-k
首先枚举列的起止范围x, y,子矩阵matrix[][x..y]可以通过部分和数组sums进行表示:
sums[i] = Σ(matrix[i][x..y])
接下来求解sums数组中和不大于k的最大子数组的和。
如果矩阵的列数远大于行数,则将枚举列变更为枚举行即可。
public int maxSumSubmatrix(int[][] matrix, int k) { int m = matrix.length, n = 0; if (m > 0) n = matrix[0].length; if (m * n == 0) return 0; int M = Math.max(m, n); int N = Math.min(m, n); int ans = Integer.MIN_VALUE; for (int x = 0; x < N; x++) { int sums[] = new int[M]; for (int y = x; y < N; y++) { TreeSet<Integer> set = new TreeSet<Integer>(); int num = 0; for (int z = 0; z < M; z++) { sums[z] += m > n ? matrix[z][y] : matrix[y][z]; num += sums[z]; if (num <= k) ans = Math.max(ans, num); Integer i = set.ceiling(num - k); if (i != null) ans = Math.max(ans, num - i); set.add(num); } } } return ans; }
https://www.hrwhisper.me/leetcode-max-sum-rectangle-no-larger-k/
朴素的思想为,枚举起始行,枚举结束行,枚举起始列,枚举终止列。。。。。O(m^2 * n^2)
这里用到一个技巧就是,进行求和时,我们可以把二维的合并成一维,然后就变为求一维的解。
比如对于矩阵:
[1, 0, 1],
[0, -2, 3]
进行起始行为0,终止行为1时,可以进行列的求和,即[1, -2, 3]中不超过k的最大值。
求和的问题解决完,还有一个是不超过k. 这里我参考了 https://leetcode.com/discuss/109705/java-binary-search-solution-time-complexity-min-max-log-max 的方法
使用了二分搜索。对于当前的和为sum,我们只需要找到一个最小的数x,使得 sum – k <=x,这样可以保证sum – x <=k。
这里需要注意,当行远大于列的时候怎么办呢?转换成列的枚举 即可。
在代码实现上,我们只需要让 m 永远小于 n即可。这样复杂度总是为O(m^2*n*log n)
int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
if (matrix.empty()) return 0;
int ans = INT_MIN,m = matrix.size(), n = matrix[0].size(),row_first=true;
if (m > n) {
swap(m, n);
row_first = false;
}
for (int ri = 0; ri < m; ri++) {
vector<int> temp(n, 0);
for (int i = ri; i >= 0; i--) {
set<int> sums;
int sum = 0;
sums.insert(sum);
for (int j = 0; j < n; j++) {
temp[j] += row_first ? matrix[i][j]: matrix[j][i];
sum += temp[j];
auto it = sums.lower_bound(sum - k);
if (it != sums.end())
ans = max(ans, sum - *it);
sums.insert(sum);
}
}
}
return ans;
}
https://leetcode.com/discuss/109705/java-binary-search-solution-time-complexity-min-max-log-max
/* first consider the situation matrix is 1D we can save every sum of 0~i(0<=i<len) and binary search previous sum to find possible result for every index, time complexity is O(NlogN). so in 2D matrix, we can sum up all values from row i to row j and create a 1D array to use 1D array solution. If col number is less than row number, we can sum up all values from col i to col j then use 1D array solution. */ public int maxSumSubmatrix(int[][] matrix, int target) { int row = matrix.length; if(row==0)return 0; int col = matrix[0].length; int m = Math.min(row,col); int n = Math.max(row,col); //indicating sum up in every row or every column boolean colIsBig = col>row; int res = Integer.MIN_VALUE; for(int i = 0;i<m;i++){ int[] array = new int[n]; // sum from row j to row i for(int j = i;j>=0;j--){ int val = 0; TreeSet<Integer> set = new TreeSet<Integer>(); set.add(0); //traverse every column/row and sum up for(int k = 0;k<n;k++){ array[k]=array[k]+(colIsBig?matrix[j][k]:matrix[k][j]); val = val + array[k]; //use TreeMap to binary search previous sum to get possible result Integer subres = set.ceiling(val-target); if(null!=subres){ res=Math.max(res,val-subres); } set.add(val); } } } return res; }
Thank you for your great answer! Can you explain why you need to set.add(0) ?? even though s(j) (j < i) may not necessarily contains 0?
Because the result can start from index 0. e.g. [1,1,5] k=3 solution is 1+1=2. So 0 in set means no need to subtract previous sum.
There is a simple version of O(n^4). The idea is to calculate every rectangle [[r1,c1], [r2,c2]], and simply pick the max area <= k. An improved version takes O(n^3logn). It borrows the idea to find max subarray with sum <= k in 1D array, and apply here: we find all rectangles bounded between r1 & r2, with columns from 0 to end. Pick a pair from tree. I remember the interviewer said there could be an even better solution, but I haven't figured that out...
public int maxSumSubmatrix(int[][] matrix, int k) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0; int rows = matrix.length, cols = matrix[0].length; int[][] areas = new int[rows][cols]; for (int r = 0; r < rows; r++) { for (int c = 0; c < cols; c++) { int area = matrix[r][c]; if (r-1 >= 0) area += areas[r-1][c]; if (c-1 >= 0) area += areas[r][c-1]; if (r-1 >= 0 && c-1 >= 0) area -= areas[r-1][c-1]; areas[r][c] = area; } } int max = Integer.MIN_VALUE; for (int r1 = 0; r1 < rows; r1++) { for (int c1 = 0; c1 < cols; c1++) { for (int r2 = r1; r2 < rows; r2++) { for (int c2 = c1; c2 < cols; c2++) { int area = areas[r2][c2]; if (r1-1 >= 0) area -= areas[r1-1][c2]; if (c1-1 >= 0) area -= areas[r2][c1-1]; if (r1-1 >= 0 && c1 -1 >= 0) area += areas[r1-1][c1-1]; if (area <= k) max = Math.max(max, area); } } } } return max; }
Solution II (O(n^3logn)
public int maxSumSubmatrix(int[][] matrix, int k) { if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0; int rows = matrix.length, cols = matrix[0].length; int[][] areas = new int[rows][cols]; for (int r = 0; r < rows; r++) { for (int c = 0; c < cols; c++) { int area = matrix[r][c]; if (r-1 >= 0) area += areas[r-1][c]; if (c-1 >= 0) area += areas[r][c-1]; if (r-1 >= 0 && c-1 >= 0) area -= areas[r-1][c-1]; areas[r][c] = area; } } int max = Integer.MIN_VALUE; for (int r1 = 0; r1 < rows; r1++) { for (int r2 = r1; r2 < rows; r2++) { TreeSet<Integer> tree = new TreeSet<>(); tree.add(0); // padding for (int c = 0; c < cols; c++) { int area = areas[r2][c]; if (r1-1 >= 0) area -= areas[r1-1][c]; Integer ceiling = tree.ceiling(area - k); if (ceiling != null) max = Math.max(max, area - ceiling); tree.add(area); } } } return max; }

https://reeestart.wordpress.com/2016/06/24/google-largest-matrix-less-than-target/
这题和Maximum Sub Matrix差不多,只不过Less than target的条件使得Kadane’s Algorithm不再适用于这道题。需要将maximum subarray改成maximum subarray less than target.
正确的方法要借助cumulative sum,
sums[j] – sums[i] <= k
sums[j] <= k + sums[i]
所以,遍历sums[]数组,对于每个sums[i],在[i+1, n-1]里面找 满足小于等于k + sums[i]中最大的。找得到就更新result,找不到就不更新。
那么问题来了,怎么找小于等于k + sums[i]中最大的?
答曰:Binary Search!
然后我就对着我的binary search code debug了快4个小时,当知道真相的那一刻眼泪都要掉下来了。
谁他妈说过cumulative sum一定是递增的?
那么问题又来了,一个个找要O(n^2) time,还有别的方法么?
“满足小于等于k + sums[i]中最大的”  ==  floor(k + sums[i])
看到floor想到什么?
Red-Black Tree啊!TreeSet啊!TreeMap啊!Balanced Binary Search Tree啊!
[Time & Space]
由于Maximum subarray less than Target需要O(nlogn) time,所以总的时间为O(n^3 * logn)
  public int maxSumSubmatrix(int[][] matrix, int k) {
    if (matrix == null || matrix.length == 0) {
      return 0;
    }
 
    int m = matrix.length;
    int n = matrix[0].length;
    int result = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {
      int r = i;
      int[] local = new int[m];
      while (r < n) {
        for (int x = 0; x < m; x++) {
 
          local[x] += matrix[x][r];
        }
 
        int localResult = maxSubArray(local, k);
        result = Math.max(result, localResult);
        r++;
      }
    }
    return result;
  }
 
  // maximum subarray less than k
  // can not use two pointer, [2, -7, 6, -1], k = 0
 
  // TreeSet, find ceiling, O(nlogn)
  private int maxSubArray(int[] nums, int k) {
    int result = Integer.MIN_VALUE;
    TreeSet<Integer> tSet = new TreeSet<>();
    int[] sums = new int[nums.length];
    sums[0] = nums[0];
    tSet.add(sums[0]);
    if (sums[0] <= k) {
      result = Math.max(result, sums[0]);
    }
    for (int i = 1; i < nums.length; i++) {
      sums[i] = sums[i - 1] + nums[i];
      if (sums[i] <= k) {
        result = Math.max(result, sums[i]);
      }
      Integer ceil = tSet.ceiling(sums[i] - k);
      if (ceil != null) {
        result = Math.max(result, sums[i] - ceil);
      }
      tSet.add(sums[i]);
    }
 
    return result;
  }
 
  // TreeSet find floor
  private int maxSubArray2(int[] nums, int k) {
    int n = nums.length;
    int result = Integer.MIN_VALUE;
    TreeSet<Integer> tSet = new TreeSet<>();
    int[] sums = new int[n];
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += nums[i];
    }
 
    for (int i = n - 1; i >= 0; i--) {
      sums[i] = sum;
      if (sums[i] <= k) {
        result = Math.max(result, sums[i]);
      }
      if (i != n - 1) {
        Integer floor = tSet.floor(sums[i] + k);
        if (floor != null) {
          result = Math.max(result, floor - sums[i]);
        }
      }
      sum -= nums[i];
      tSet.add(sums[i]);
    }
    return result;
  }


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Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

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