Given two integers N and M which denote the number of persons of Type1 and Type2 respectively. The task is to find the maximum number of teams that can be formed with these two types of persons. A team can contain either 2 persons of Type1 and 1 person of Type2 or 1 person of Type1 and 2 persons of Type2.
Input: N = 2, M = 6
(Type1, Type2, Type2) and (Type1, Type2, Type2) are the two possible teams.
Input: N = 4, M = 5
Approach: A greedy approach is to choose 2 persons from the group which has more members and 1 person from the group with lesser members and update the count of persons in each of the group accordingly. Termination condition will be when no more teams can be formed.
Below is the implementation of the above approach:
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- Greatest number less than equal to B that can be formed from the digits of A
- Check if the large number formed is divisible by 41 or not
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