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    <p>and that's how the sample input looks? Sorted by diagonals? That's an interesting sort, to be sure. </p> <p>Since the following row may have a value that's lower than any value on this row, you can't assume anything in particular about a given row of data. </p> <p>I would (if asked to do this over a large input) read the matrix into a list-struct that took the data as one pair of a tuple, and the mxn coord as the part of the tuple, and then quicksort the matrix once, then find it by value. </p> <p>Alternately, if the value of each individual location is unique, toss the MxN data into a dictionary keyed on the value, then jump to the dictionary entry of the MxN based on the key of the input (or the hash of the key of the input).</p> <p>EDIT:</p> <p>Notice that the answer I give above is valid if you're going to look through the matrix more than once. If you only need to parse it once, then this is as fast as you can do it:</p> <pre><code>for (int i = 0; i&lt;M; i++) for (int j=0; j&lt;N; j++) if (mat[i][j] == value) return tuple(i,j); </code></pre> <hr> <p>Apparently my comment on the question should go down here too :|</p> <blockquote> <p>@sagar but that's not the example given by the professor. otherwise he had the fastest method above (check the end of the row first, then proceed) additionally, checking the end of the middlest row first would be faster, a bit of a binary search.</p> </blockquote> <p>Checking the end of each row (and starting on the end of the middle row) to find a number higher than the checked for number on an in memory array would be fastest, then doing a binary search on each matching row till you find it.</p>
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    1. POMost efficient way to search a sorted matrix?
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    1. COwell the diagonals do appear to be sorted, didn't notice that before. but the sort is not by diagonals, it is simpler. every row by itself is sorted and every column by itself is sorted. but like you said reading a row doesn't give any information about the next row.
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    2. COnor a column about the next column. You can make no assumptions about the speed of finding an element, without iterating over every one. I'm trying to look for the fastest possible method of finding a value. An O(n) seems like the fastest, but if you're going to do what I suggested, that's only valid if you're looking for lots of the values. Otherwise, if you just parse it once, then it's faster to just for loop it twice on the MxN with an early abort on found.
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    3. CObut as soon as you need to find two values, mine way given in the answer is faster than a double for loop
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