## Mentor Graphics Interview Question for MTSs

• 0

Country: India
Interview Type: In-Person

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1)start from top right corner
2) until curr_elem==elem_to_search || you go out of matrix, go down if curr_elem<elem_to_search else go left

complexity= O(m+n)

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thr is a specific name of such kinds of matrices....ny idea nyone?

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yup , there is a name it's called Young's tableau.

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0

Yes

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0
of 0 vote

Do u mean that rows and columns are sorted in themselves independently ? ie.
3 6 7 12
7 9 10 13
9 12 13 14

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0
of 0 vote

number of solutions are possible :
1) if min(m,n)==1 use binary search.

2) if min(m,n)<<small linear search in zig-zag manner starting at top right corner. O(max(m,n))

3) if both m and n large use 2D binary-search O(log(n)*log(m))

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0
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``````public static void findInArray(int[][] inpArr, int num) {
int count = 1;
boolean flag = true;
int i = 0;
int j = (inpArr[0].length)-1;
int currentNum = inpArr[i][j];
while(num!=currentNum) {
count++;
if(num>currentNum) {
i++;
} else {
j--;
}
try {
currentNum = inpArr[i][j];
} catch(ArrayIndexOutOfBoundsException e) {
flag = false;
break;
}
}
if(flag) System.out.println(num + "fount at (" + i +","+ j + ") position in " + count + " comparision(s).");
}``````

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0

While this solution is correct, it does not make use of sorted columns and rows.
Following is the algo:
1. Look along diagonal for element greater than element to be found. Let's call this point (row, col) as pivot.
2. The pivot point cuts the array into 4 quadrant. Our element can only be present in the quadrant that is
2(a) above and towards right
2(b) below and towards left of this pivot point.
3. Apply this recursively to get the ans.

Name:

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