Given n instance of item A and m instance of item B. Find the maximum number of groups of size 3 that can be formed using these items such that all groups contain items of both types, i.e., a group should not have either all items of type A or all items of type B.
Total number of items of type A in the formed groups must be less than or equal to n.
Total number of items of type B in the formed groups must be less than or equal to m.
Input : n = 2 and m = 6. Output : 2 In group 1, one item of type A and two items of type B. Similarly, in the group 2, one item of type A and two items of type B. We have used 2 (<= n) items of type A and 4 (<= m) items of type B. Input : n = 4 and m = 5. Output : 3 In group 1, one item of type A and two items of type B. In group 2, one item of type B and two items of type A. In group 3, one item of type A and two items of type B. We have used 4 (<= n) items of type A and 5 (<= 5) items of type B.
1. There will be n groups possible if m >= 2n. Or there will be m groups possible, if n >= 2m.
2. Suppose n = 3 and m = 3, so one instance of item A will make a group with the two instance of item B and one instance of item B will make a group with the two instance of item A. So, maximum two groups are possible. So find the total number of such conditions with given n and m by dividing m and m by 3. After this, there can be 0, 1, 2 instances of each type can be left. For finding the number of groups for the left instances:
a) If n = 0 or m = 0, 0 group is possible.
b) If n + m >= 3, only 1 group is possible.
Algorithm for solving this problem:
1. If n >= 2m, maximum number of groups = n.
2. Else if m >= 2n, maximum number of groups = m.
3. Else If (m + n) % 3 == 0, maximum number of group = (m + n)/3;
4. Else maximum number of group = (n + m)/3. And set n = n%3 and m = m%3.
a) If n != 0 && m != 0 && (n + m) >= 3, add one to maximum number of groups.
Below is implementation of the above idea :
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