最大的按行和按列排序的子矩阵
给定一个N * M矩阵mat[][] ，任务是找到面积最大的矩形子矩阵，使得子矩阵的每一列和每一行都严格递增。
例子：
Input: mat[][] =
{{1, 2, 3},
{4, 5, 6},
{1, 2, 3}}
Output: 6
Largest sub-matrix will be {{1, 2, 3}, {4, 5, 6}}.
Number of elements in this sub-matrix = 6.
Input: mat[][] =
{{1, 2, 3},
{4, 5, 3},
{1, 2, 3}}
Output: 4
The largest sub-matrix will be
{{1, 2}, {4, 5}}
方法：有很多方法可以解决这个问题，从 O(N 3 * M 3 ) 到 O(N * M)。在本文中，将讨论使用堆栈的具有 O(N * M) 时间复杂度的方法。
在继续之前，建议解决此问题。问题。
让我们尝试广泛地理解该方法，然后将讨论该算法。对于矩阵的每一列，尝试找到在该列具有左边缘的最大的按行和按列排序的子矩阵。要执行相同的操作，请创建一个矩阵pre[][]其中pre[i][j]将存储从数组arr[i]的索引j开始的最长递增子数组的长度。
现在使用这个矩阵，对于每一列j ，找到最长的行和列排序数组的长度。要处理一列，将需要数组pre[][j]的所有递增子段。使用两指针技术也可以找到相同的结果。在这些子段中的每一个中，只需将逐行增加的子段视为条形，即可找到直方图下的最大区域。
- 为每一行“i”创建一个前缀和数组，该数组存储在该行的每一列“j”处结束的最大递增子数组的长度。
- 一旦我们有了这个数组，对于每一列'j'。
- 初始化 'i' 等于 0。
- 在“i”小于“N”时在“i”上运行循环
- 初始化 'k' 等于 i+1。
- 当 k 小于 N 且 arr[k][j] 大于 arr[k-1][j] 时，增加 k。
- 将子数组 pre[i][j] 上的直方图问题应用到 pre[k-1][j] 上，找到它下面的最大区域。让我们将此值称为“V”。将最终答案更新为 ans = max(ans, val)。
- 更新 'i' 等于 k-1。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the largest
// area under a histogram
int histo(vector q)
{
// Stack
stack q1;
// Length of the vector
int n = q.size();
// Function to store the next smaller
// and previous smaller index
int pre_smaller[q.size()];
int next_smaller[q.size()];
// Finding the next smaller
for (int i = 0; i < n; i++)
pre_smaller[i] = -1, next_smaller[i] = n;
for (int i = 0; i < n; i++) {
while (q1.size() and q[q1.top()] > q[i]) {
next_smaller[q1.top()] = i;
q1.pop();
}
q1.push(i);
}
// Finding the previous smaller element
while (q1.size())
q1.pop();
for (int i = n - 1; i >= 0; i--) {
while (q1.size() and q[q1.top()] > q[i]) {
pre_smaller[q1.top()] = i;
q1.pop();
}
q1.push(i);
}
// To store the final answer
int ans = 0;
// Finding the final answer
for (int i = 0; i < n; i++)
ans = max(ans, (next_smaller[i]
- pre_smaller[i] - 1)
* q[i]);
// Returning the final answer
return ans;
}
// Function to return the largest area
// for the required submatrix
int findLargest(vector > arr)
{
// n and m store the number of
// rows and columns respectively
int n = arr.size();
int m = arr[0].size();
// To store the prefix_sum
int pre[n][m];
// To store the final answer
int ans = 0;
// Loop to create the prefix-sum
// using two pointers
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
if (j == 0) {
pre[i][j] = 1;
continue;
}
if (arr[i][j] > arr[i][j - 1])
pre[i][j] = pre[i][j - 1] + 1;
else
pre[i][j] = 1;
}
// For each column run the loop
for (int j = 0; j < m; j++) {
// Find the largest row-wise sorted arrays
for (int i = 0; i < n; i++) {
int k = i + 1;
vector q;
q.push_back(pre[i][j]);
while (k < n and arr[k] > arr[k - 1])
q.push_back(pre[k][j]), k++;
// Applying the largest area
// under the histogram
ans = max(ans, histo(q));
i = k - 1;
}
}
// Return the final answer
return ans;
}
// Driver code
int main()
{
vector > arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 1, 2, 3 } };
cout << findLargest(arr);
return 0;
} Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to return the largest
// area under a histogram
static int histo(ArrayList q)
{
// Stack
Stack q1 = new Stack();
// Length of the vector
int n = q.size();
// Function to store the next smaller
// and previous smaller index
int[] pre_smaller = new int[q.size()];
int[] next_smaller = new int[q.size()];
// Finding the next smaller
for (int i = 0; i < n; i++)
{
pre_smaller[i] = -1;
next_smaller[i] = n;
}
for (int i = 0; i < n; i++)
{
while (q1.size() > 0 && q.get(q1.peek()) > q.get(i))
{
next_smaller[q1.peek()] = i;
q1.pop();
}
q1.push(i);
}
// Finding the previous smaller element
while (q1.size() > 0)
{
q1.pop();
}
for (int i = n - 1; i >= 0; i--)
{
while (q1.size() > 0 && q.get(q1.peek()) > q.get(i))
{
pre_smaller[q1.peek()] = i;
q1.pop();
}
q1.push(i);
}
// To store the final answer
int ans = 0;
// Finding the final answer
for (int i = 0; i < n; i++)
{
ans = Math.max(ans, (next_smaller[i] -
pre_smaller[i] - 1) *
q.get(i));
}
// Returning the final answer
return ans;
}
// Function to return the largest area
// for the required submatrix
static int findLargest(ArrayList> arr)
{
// n and m store the number of
// rows and columns respectively
int n = arr.size();
int m = arr.get(0).size();
// To store the prefix_sum
int[][] pre=new int[n][m];
// To store the final answer
int ans = 0;
// Loop to create the prefix-sum
// using two pointers
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (j == 0)
{
pre[i][j] = 1;
continue;
}
if(arr.get(i).get(j) > arr.get(i).get(j - 1))
{
pre[i][j] = pre[i][j - 1] + 1;
}
else
{
pre[i][j] = 1;
}
}
}
// For each column run the loop
for (int j = 0; j < m; j++)
{
// Find the largest row-wise sorted arrays
for (int i = 0; i < n; i++)
{
int k = i + 1;
ArrayList q = new ArrayList();
q.add(pre[i][j]);
while (k < n && arr.get(k).get(0) > arr.get(k - 1).get(0))
{
q.add(pre[k][j]);
k++;
}
// Applying the largest area
// under the histogram
ans = Math.max(ans, histo(q));
i = k - 1;
}
}
// Return the final answer
return ans;
}
// Driver code
public static void main (String[] args)
{
ArrayList> arr = new ArrayList>();
arr.add(new ArrayList(Arrays.asList(1, 2, 3 )));
arr.add(new ArrayList(Arrays.asList(4, 5, 6 )));
arr.add(new ArrayList(Arrays.asList(1, 2, 3 )));
System.out.println(findLargest(arr));
}
}
// This code is contributed by avanitrachhadiya2155 Python3
# Python3 implementation of the approach
# Function to return the largest
# area under a histogram
def histo(q):
# Stack
q1 = []
# Length of the vector
n = len(q)
# Function to store the next smaller
# and previous smaller index
pre_smaller = [0 for i in range(len(q))]
next_smaller = [0 for i in range(len(q))]
# Finding the next smaller
for i in range(n):
pre_smaller[i] = -1
next_smaller[i] = n
for i in range(n):
while (len(q1) > 0 and q[q1[-1]] > q[i]):
next_smaller[q1[-1]] = i
del q1[-1]
q1.append(i)
# Finding the previous smaller element
while (len(q1) > 0):
del q1[-1]
for i in range(n - 1, -1, -1):
while (len(q1) > 0 and q[q1[-1]] > q[i]):
pre_smaller[q1[-1]] = i
del q1[-1]
q1.append(i)
# To store the final answer
ans = 0
# Finding the final answer
for i in range(n):
ans = max(ans, (next_smaller[i]- pre_smaller[i] - 1)* q[i])
# Returning the final answer
return ans
# Function to return the largest area
# for the required submatrix
def findLargest(arr):
# n and m store the number of
# rows and columns respectively
n = len(arr)
m = len(arr[0])
# To store the prefix_sum
pre = [[0 for i in range(m)] for i in range(n)]
# To store the final answer
ans = 0
# Loop to create the prefix-sum
# using two pointers
for i in range(n):
for j in range(m):
if (j == 0):
pre[i][j] = 1
continue
if (arr[i][j] > arr[i][j - 1]):
pre[i][j] = pre[i][j - 1] + 1
else:
pre[i][j] = 1
# For each column run the loop
for j in range(m):
# Find the largest row-wise sorted arrays
for i in range(n):
k = i + 1
q = []
q.append(pre[i][j])
while (k < n and arr[k] > arr[k - 1]):
q.append(pre[k][j])
k += 1
# Applying the largest area
# under the histogram
ans = max(ans, histo(q))
i = k - 1
# Return the final answer
return ans
# Driver code
arr = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 1, 2, 3 ] ]
print(findLargest(arr))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
// Function to return the largest
// area under a histogram
static int histo(List q)
{
// Stack
Stack q1 = new Stack();
// Length of the vector
int n = q.Count;
// Function to store the next smaller
// and previous smaller index
int[] pre_smaller = new int[q.Count];
int[] next_smaller = new int[q.Count];
// Finding the next smaller
for(int i = 0; i < n; i++)
{
pre_smaller[i] = -1;
next_smaller[i] = n;
}
for(int i = 0; i < n; i++)
{
while (q1.Count > 0 && q[q1.Peek()] > q[i])
{
next_smaller[q1.Peek()] = i;
q1.Pop();
}
q1.Push(i);
}
// Finding the previous smaller element
while (q1.Count > 0)
{
q1.Pop();
}
for(int i = n - 1; i >= 0; i--)
{
while (q1.Count > 0 && q[q1.Peek()] > q[i])
{
pre_smaller[q1.Peek()] = i;
q1.Pop();
}
q1.Push(i);
}
// To store the final answer
int ans = 0;
// Finding the final answer
for(int i = 0; i < n; i++)
{
ans = Math.Max(ans, (next_smaller[i] -
pre_smaller[i] - 1) * q[i]);
}
// Returning the
// final answer
return ans;
}
// Function to return the largest area
// for the required submatrix
static int findLargest(List> arr)
{
// n and m store the number of
// rows and columns respectively
int n = arr.Count;
int m = arr[0].Count;
// To store the prefix_sum
int[,] pre = new int[n, m];
// To store the final answer
int ans = 0;
// Loop to create the prefix-sum
// using two pointers
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if (j == 0)
{
pre[i, j] = 1;
continue;
}
if (arr[i][j] > arr[i][j - 1])
{
pre[i, j] = pre[i,j - 1] + 1;
}
else
{
pre[i, j] = 1;
}
}
}
// For each column run the loop
for(int j = 0; j < m; j++)
{
// Find the largest row-wise sorted arrays
for(int i = 0; i < n; i++)
{
int k = i + 1;
List q = new List();
q.Add(pre[i, j]);
while(k < n && arr[k][0] > arr[k - 1][0])
{
q.Add(pre[k, j]);
k++;
}
// Applying the largest area
// under the histogram
ans = Math.Max(ans, histo(q));
i = k - 1;
}
}
// Return the final answer
return ans;
}
// Driver code
static public void Main()
{
List> arr = new List>();
arr.Add(new List(){1, 2, 3});
arr.Add(new List(){4, 5, 6 });
arr.Add(new List(){1, 2, 3});
Console.WriteLine(findLargest(arr));
}
}
// This code is contributed by rag2127 Javascript
输出：
6时间复杂度： O(N 2 )
辅助空间：O(N 2 )。