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You are given a collection of positive integers and a target number of groups k. Your task is to determine whether it is possible to distribute all the integers from the collection into exactly k non-empty groups such that every group has the same total sum.
Each integer must be placed in exactly one group, and no integer can be left unassigned or used in multiple groups. The order of integers within a group does not matter, and the groups themselves are not ordered.
This problem requires you to explore different ways of assigning integers to groups and verify if a valid assignment exists. The challenge lies in the combinatorial explosion of possibilities—each integer can potentially go into any of the k groups, leading to an exponential search space.
Before embarking on an exhaustive search, several quick checks can immediately determine if a valid partition is impossible:
k, partitioning is impossible. Each group must sum to total / k.The backtracking approach attempts to build groups one at a time. For each integer, try placing it in one of the existing groups (if the group sum would not exceed the target) or start a new group. If a configuration leads to a dead end, backtrack and try alternative placements.
The problem elegantly bridges greedy intuitions with rigorous algorithmic search, making it a classic example of optimization through intelligent backtracking.
nums = [4,3,2,3,5,2,1]
k = 4trueThe total sum is 20, so each group must sum to 5. One valid distribution is: {5}, {4,1}, {3,2}, {3,2}. Each group sums to exactly 5, confirming the partition is possible.
nums = [1,2,3,4]
k = 3falseThe total sum is 10, which is not divisible by 3. Since 10/3 is not an integer, we cannot form 3 groups with equal integer sums. Hence, partitioning is impossible.
nums = [2,2,2,2,2,2]
k = 3trueThe total sum is 12, so each group must sum to 4. One valid distribution is: {2,2}, {2,2}, {2,2}. Each group contains two 2s, summing to 4.
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