CareerCup

Less than O(n^2) makes no sense. Go read up on the definition of BigOh.

As to an approach: here is a greedy approach: Pick a random element and pick the neighbour which is the smallest and less than it. (If there is no such neigbour, you found it). Repeat the process with the neighbour.

You will end up at some 'sink'. Not sure if the time complexity would be sub-quadratic, perhaps we can prove it...

Less than o(n^2) means o(nlogn) or in between both??

there is no answer.

#include <cstdio>

const int N = 5;
const int M = 5;

int a[N][M] = {1, 12, 3, 1, -23,  
			   7, 9, 8, 5, 6,
			   4, 5, 6, -1, 77,
			   7, 0, 35, -2, -4,
			   6, 83, 1, 7, -6};

int direct[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

void findLocalMin(int& resRow, int& resCol)
{	

	int curRow = N / 2;

	int curCol = 0;
	int curMin = a[curRow][curCol];

	for(int i = 0; i < M; i++)
	{
		if (a[curRow][i] < curMin)
		{
			curMin = a[curRow][i];
			curCol = i;
		}
	}

	bool flag = false;

	while(true)
	{
		flag = true;
		for (int i = 0; i < 4; i++)
		{
			int tempRow = curRow + direct[i][0];
			int tempCol = curCol + direct[i][1];

			if (tempRow < N && 
				tempCol < N && 
				a[curRow][curCol] > a[tempRow][tempCol])
			{
				curMin = a[tempRow][tempCol];
				curRow = tempRow; curCol = tempCol;
				flag = false;
				break;
			}
		}

		if (flag)
		{
			resRow = curRow; 
			resCol = curCol;
			return;
		}		
	}
};

int main()
{
	int resRow, resCol;
	findLocalMin(resRow, resCol);
	printf("%d", a[resRow][resCol]);
	return 0;
}

here we have few points to consider we are not checking the elemnts in the boundaries
1)first we find the rows to be checkd
2)take an array of size number of rows which contains the info. as which rows are checkd

here is the idea with rough code..it may have mistakes suggest me if am wrong anywhere
#define rowno 5
static int count=0;//counts number of such elemnts
int d[r0wno];//array to keep track of rows checked
for(int i=0;i<rowno;i++)
d[i]=0;


findrows(int a[][],int start,int end)
{
int mid=(start+end)/2;
check(a,mid);
if((mid+1)!=(rowno-1) && a[mid+1]!=1)//checking if its last row
{
a[mid+1]=1;
check(a,mid+1);
}
if((mid-1)!=0 && a[mid-1]!=1)//checking if its first row
{
a[mid-1]=1;
check(a,mid-1);
}
findrows(a,start,mid-1);
findrows(a,mid+1,end);
}

check(a[][],int ind)
{
//start from 2nd column because first is boundary
for(int j=1;j<rowno-2;j++)
{
if(a[ind][j]<a[ind][j-1] && a[ind][j]<a[ind[j+1])
{
if(a[ind][j]<a[ind-1][j] && a[ind][j]<a[ind+1][j])
{
count++;
printf("found element %d",a[ind][j]);
}
}
}
}

sry there a few mistakes in my previous code..modified code is
here we have few points to consider we are not checking the elemnts in the boundaries
1)first we find the rows to be checkd
2)take an array of size number of rows which contains the info. as which rows are checkd

here is the idea with rough code..it may have mistakes suggest me if am wrong anywhere
#define rowno 5
static int count=0;//counts number of such elemnts
int d[r0wno];//array to keep track of rows checked
for(int i=0;i<rowno;i++)
d[i]=0;

findrows(int a[][],int start,int end)
{
int mid=(start+end)/2;
check(a,mid);
if((mid+1)!=(rowno-1) && d[mid+1]!=1)//checking if its last row
{
d[mid+1]=1;
check(a,mid+1);
}
if((mid-1)!=0 && dmid-1]!=1)//checking if its first row
{
d[mid-1]=1;
check(a,mid-1);
}
findrows(a,start,mid-1);
findrows(a,mid+1,end);
}

check(a[][],int ind)
{
//start from 2nd column because first is boundary
for(int j=1;j<rowno-2;j++)
{
if(a[ind][j]<a[ind][j-1] && a[ind][j]<a[ind[j+1])
{
if(a[ind][j]<a[ind-1][j] && a[ind][j]<a[ind+1][j])
{
count++;
printf("found element %d",a[ind][j]);
}
}
}
}

//call findrows(a[][],0,rowno-1);

int main()
{
int a[5][5]=
{
10, 2, 4, 7, 11,
3, 5, 8, 12, 17,
6, 9, 13, 18, 22,
10, 14, 19, 23, 25,
15, 20, 24, 26, 27
};
int i=0,j;
while(1)
{
for(j=0;j<=4;j++)
{
if(i==0 || j==0)
{
if(a[i][j]<a[i+1][j] && a[i][j]<a[i][j+1])
{
printf("i=%d j=%d\n",i,j);
getch();
}
}
else
{
if(a[i][j]<a[i+1][j] && a[i][j]<a[i][j+1] && a[i][j]<a[i-1][j] && a[i][j]< a[i][j-1])
{
printf("i=%d j=%d\n",i,j);
getch();
}
}
}
i++;
if(i==5)
break;
}
}

This can be done in O(n) time. If we divide the matrix into 4 equal sub-matrices, one of them must contain such (i, j). We can identify that sub-matrix by finding the smallest elements of the middle row and middle column. If this element satisfies the condition, then output it. Otherwise, it has a smaller neighbor. We continue to detect on the sub-matrix that contain this smaller neighbor. The running time is 2n + n + n/2 ... = 4n.

public static int[] findLocalMimMatrix(int[][] matrix, int startRow, int endRow){
		int[] ret = new int[2];
		int midRow = startRow + (endRow - startRow) / 2;
		int rowMin = matrix[midRow][0];
		ret[0] = midRow;
		ret[1] = 0;
		// Get the minimum value of row (startRow + endRow ) / 2
		for(int i = 1; i < matrix[midRow].length; i ++){
			if(matrix[midRow][i] < rowMin){
				rowMin = matrix[midRow][i];
				ret[0] = midRow;
				ret[1] = i;
			}
		}
		int upNbr = Integer.MAX_VALUE, downNbr = Integer.MAX_VALUE;
		if(midRow - 1 >= startRow){
			upNbr = matrix[midRow - 1][ret[1]];
		}
		
		if(midRow + 1 <= endRow){
			downNbr = matrix[midRow + 1][ret[1]];
		}
		
		// Find a local minimum
		if(rowMin <= upNbr && rowMin <= downNbr){
			return ret;
		}else{
			// Shrink the matrix into half.
			if(upNbr >= downNbr){
				ret = findLocalMimMatrix(matrix, midRow + 1, endRow);
			}else{
				ret = findLocalMimMatrix(matrix, startRow, midRow - 1);
			}
			return ret;
		}
	}

The time complexity is O(N*lgN)

We know that we dont have to check the outer elements of the matrix.
Thus, we are left with (N-2) rows as well as column.
Hence, the time complexity is (N-2)^2