You are given N jobs with their start and end time mentioned.These jobs may have their timings overlapped.You have to suggest an algorithm such that maximum number of jobs can be performed in that given time interval.
Arrange the activities in increasing order of their finish times. Select the first activity. If any activity's starting time is greater than the finish time of last selected activity, select that activity...otherwise discard it and so on..
Given a Binary Tree, find vertical sum of the nodes that are in same vertical line. Print all sums through different vertical lines. Examples: 1 / \ 2 3 / \ / \ 4 5 6 7 The tree has 5 vertical lines Vertical-Line-1 has only one node 4 => vertical sum is 4 Vertical-Line-2: has only one node 2=> vertical sum is 2 Vertical-Line-3: has three nodes: 1,5,6 => vertical sum is 1+5+6 = 12 Vertical-Line-4: has only one node 3 => vertical sum is 3 Vertical-Line-5: has only one node 7 => vertical sum is 7 So expected output is 4, 2, 12, 3 and 7
Arrange the activities in increasing order of their finish times. Select the first activity. If any activity's starting time is greater than the finish time of last selected activity, select that activity...otherwise discard it and so on..
ReplyDelete@Anonymous yes your approach is correct.can you please post a code for others..:)
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