# Difference between revisions of "TADM2E 4.35"

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− | + | <math>O(n+m)</math> is necessary and sufficient. | |

Lower bound comes from potentially independent values along second diagonal -- | Lower bound comes from potentially independent values along second diagonal -- | ||

upper bound comes from observing that we can eliminate either a row or a | upper bound comes from observing that we can eliminate either a row or a | ||

Line 9: | Line 9: | ||

Point* findPosition(int key) { | Point* findPosition(int key) { | ||

int row = 0, col = m-1; | int row = 0, col = m-1; | ||

− | while (row | + | while (row < n && col >= 0) { |

if (M[row][col] == key) { | if (M[row][col] == key) { | ||

return new Point(row,col); | return new Point(row,col); | ||

} | } | ||

− | else if (M[row][col] | + | else if (M[row][col] > key) { |

col--; | col--; | ||

} | } |

## Revision as of 18:23, 11 September 2014

$ O(n+m) $ is necessary and sufficient. Lower bound comes from potentially independent values along second diagonal -- upper bound comes from observing that we can eliminate either a row or a column in each comparison if we start from the lower left corner and walk up or left.

Having M[0..n-1][0..m-1] and a struct Point {int x, int y}, we could have the following solution:

Point* findPosition(int key) {

int row = 0, col = m-1; while (row < n && col >= 0) { if (M[row][col] == key) { return new Point(row,col); } else if (M[row][col] > key) { col--; } else row++; } return NULL;

}