Next Greater Element

Next Greater Element

Given an array, print the Next Greater Element (NGE) for every element. The Next greater Element for an element x is the first greater element on the right side of x in array. Elements for which no greater element exist, consider next greater element as -1.

Solution:

1) Push the first element to stack.
2) Pick rest of the elements one by one and follow following steps in loop.
....a) Mark the current element as next.
....b) If stack is not empty, then pop an element from stack and compare it with next.
....c) If next is greater than the popped element, then next is the next greater element fot the popped element.
....d) Keep poppoing from the stack while the popped element is smaller than next. next becomes the next greater element for all such popped elements
....g) If next is smaller than the popped element, then push the popped element back.
3) After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.

void printNGE(int arr[], int n)

{

int i = 0;

struct stack s;

s.top = -1;

int element, next;

/* push the first element to stack */

push(&s, arr[0]);

// iterate for rest of the elements

for (i=1; i

{

next = arr[i];

if (isEmpty(&s) == false)

{

// if stack is not empty, then pop an element from stack

element = pop(&s);

/* If the popped element is smaller than next, then

a) print the pair

b) keep popping while elements are smaller and

stack is not empty */

while (element < next)

{

printf("\n %d --> %d", element, next);

if(isEmpty(&s) == true)

break;

element = pop(&s);

}

/* If element is greater than next, then push

the element back */

if (element > next)

push(&s, element);

}

/* push next to stack so that we can find

next greater for it */

push(&s, next);

}

/* After iterating over the loop, the remaining

elements in stack do not have the next greater

element, so print -1 for them */

while(isEmpty(&s) == false)

{

element = pop(&s);

next = -1;

printf("\n %d --> %d", element, next);

}

}

A Boolean Matrix Question

A Boolean Matrix Question

Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1.
if a[i][j]=1
then fill a[i][k]=1 where k 0 to N
and fill a[k][j]=1 where k 0 to M

Method 1 (Use two temporary arrays)
1) Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] as 0.
2) Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true.
3) Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.

void modifyMatrix(bool mat[R][C])

{

bool row[R]={0};

bool col[C]={0};

int i, j;

/* Store the rows and columns to be marked as 1 in row[] and col[]

arrays respectively */

for (i = 0; i < R; i++)

{

for (j = 0; j < C; j++)

{

if (mat[i][j] == 1)

{

row[i] = 1;

col[j] = 1;

}

}

}

/* Modify the input matrix mat[] using the above constructed row[] and

col[] arrays */

for (i = 0; i < R; i++)

{

for (j = 0; j < C; j++)

{

if ( row[i] == 1 || col[j] == 1 )

{

mat[i][j] = 1;

}

}

}

}

Method 2 (A Space Optimized Version of Method 1)
This method is a space optimized version of above method 1. This method uses the first row and first column of the input matrix in place of the auxiliary arrays row[] and col[] of method 1. So what we do is: first take care of first row and column and store the info about these two in two flag variables rowFlag and colFlag. Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1).

1) Scan the first row and set a variable rowFlag to indicate whether we need to set all 1s in first row or not.
2) Scan the first column and set a variable colFlag to indicate whether we need to set all 1s in first column or not.
3) Use first row and first column as the auxiliary arrays row[] and col[] respectively, consider the matrix as submatrix starting from second row and second column and apply method 1.
4) Finally, using rowFlag and colFlag, update first row and first column if needed.

Time Complexity: O(M*N)
Auxiliary Space: O(1)

void modifyMatrix(bool mat[R][C])

{

bool rowFirst;

bool colFirst;

int i, j;

for (i = 0; i < C; i++)

{

if(mat[0][i]){ rowFirst=1; break;


}

for (i = 0; i < R; i++)

{

if(mat[i][o]) { colFirst=1 ; break;}


}

for (i = 1; i < R; i++)

{

for (j = 1; j < C; j++)

{

if (mat[i][j] == 1)

{

mat[i][0] = 1;

mat[0][j] = 1;

}

}

}

for (i = 1; i < R; i++)

{

for (j = 1; j < C; j++)

{

if ( mat[i][0] == 1 || mat[0][j] == 1 )

{

mat[i][j] = 1;

}

}

}

if(rowFirst)
for (i = 0; i < R; i++)
mat[0][i]=1;
if(colFirst)
for (i = 0; i < C; i++)
mat[i][0]=1;


}