Solution

Approach #1 Using HashMap [Accepted]

In this approach, we compare every string in $$list1$$ and $$list2$$ by traversing over the whole list $$list2$$ for every string chosen from $$list1$$ . We make use of a hashmap $$map$$ , which contains elements of the form $$(sum : list_{sum})$$ . Here, $$sum$$ refers to the sum of indices of matching elements and $$list_{sum}$$ refers to the list of matching strings whose indices\' sum equals $$sum$$ .

Thus, while doing the comparisons, whenever a match between a string at $$i^{th}$$ index of $$list1$$ and $$j^{th}$$ index of $$list2$$ is found, we make an entry in the $$map$$ corresponding to the sum $$i + j$$ , if this entry isn\'t already present. If an entry with this sum already exists, we need to keep a track of all the strings which lead to the same index sum. Thus, we append the current string to the list of strings corresponding to sum $$i + j$$ .

At the end, we traverse over the keys of the $$map$$ and find out the list of strings corresponding to the key reprsenting the minimum sum.

Complexity Analysis

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  Time complexity : $$O(l_1*l_2*x)$$ . Every item of $$list1$$ is compared with all the items of $$list2$$ . $$l_1$$ and $$l_2$$ are the lengths of $$list1$$ and $$list2$$ respectively. And $$x$$ refers to average string length.

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  Space complexity : $$O(l_1*l_2*x)$$ . In worst case all items of $$list1$$ and $$list2$$ are same. In that case, hashmap size grows upto $$l_1*l_2*x$$ , where $$x$$ refers to average string length.

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Approach #2 Without Using HashMap [Accepted]

Algorithm

Another method could be to traverse over the various $$sum$$ (index sum) values and determine if any such string exists in $$list1$$ and $$list2$$ such that the sum of its indices in the two lists equals $$sum$$ .

Now, we know that the value of index sum, $$sum$$ could range from 0 to $$m + n - 1$$ . Here, $$m$$ and $$n$$ refer to the length of lists $$list1$$ and $$list2$$ respectively. Thus, we choose every value of $$sum$$ in ascending order. For every $$sum$$ chosen, we iterate over $$list1$$ . Suppose, currently the string at $$i^{th}$$ index in $$list1$$ is being considered. Now, in order for the index sum $$sum$$ to be the one corresponding to matching strings in $$list1$$ and $$list2$$ , the string at index $$j$$ in $$list2$$ should match the string at index $$i$$ in $$list1$$ , such that $$sum = i + j$$ .

Or, stating in other terms, the string at index $$j$$ in $$list2$$ should be equal to the string at index $$i$$ in $$list1$$ , such that $$j = sum - i$$ . Thus, for a particular $$sum$$ and $$i$$ (from $$list1$$ ), we can directly determine that we need to check the element at index $$j= sum - i$$ in $$list2$$ , instead of traversing over the whole $$list2$$ .

Doing such checks/comparisons, iterate over all the indices of $$list1$$ for every $$sum$$ value chosen. Whenver a match occurs between $$list1$$ and $$list2$$ , we put the matching string in a list $$res$$ .

We do the same process of checking the strings for all the values of $$sum$$ in ascending order. After completing every iteration over $$list1$$ for a particular $$sum$$ , we check if the $$res$$ list is empty or not. If it is empty, we need to continue the process with the next $$sum$$ value considered. If not, the current $$res$$ gives the required list with minimum index sum. This is because we are already considering the index sum values in ascending order. So, the first list to be found is the required resultant list.

The following example depicts the process:

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Complexity Analysis

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  Time complexity : $$O((l_1+l_2)^2*x)$$ . There are two nested loops upto $$l_1+l_2$$ and string comparison takes $$x$$ time. Here, $$x$$ refers to the average string length.

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  Space complexity : $$O(r*x)$$ . $$res$$ list is used to store the result. Assuming $$r$$ is the length of $$res$$ .

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Approach #3 Using HashMap (linear) [Accepted]

We make use of a HashMap to solve the given problem in a different way in this approach. Firstly, we traverse over the whole $$list1$$ and create an entry for each element of $$list1$$ in a HashMap $$map$$ , of the form $$(list[i], i)$$ . Here, $$i$$ refers to the index of the $$i^{th}$$ element, and $$list[i]$$ is the $$i^{th}$$ element itself. Thus, we create a mapping from the elements of $$list1$$ to their indices.

Now, we traverse over $$list2$$ . For every element , $$list2[j]$$ , of $$list2$$ encountered, we check if the same element already exists as a key in the $$map$$ . If so, it means that the element exists in both $$list1$$ and $$list2$$ . Thus, we find out the sum of indices corresponding to this element in the two lists, given by $$sum = map.get(list[j]) + j$$ . If this $$sum$$ is lesser than the minimum sum obtained till now, we update the resultant list to be returned, $$res$$ , with the element $$list2[j]$$ as the only entry in it.

If the $$sum$$ is equal to the minimum sum obtained till now, we put an extra entry corresponding to the element $$list2[j]$$ in the $$res$$ list.

Below code is inspired by @cloud.runner

Complexity Analysis

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  Time complexity : $$O(l_1+l_2)$$ . Every item of $$list2$$ is checked in a map of $$list1$$ . $$l_1$$ and $$l_2$$ are the lengths of $$list1$$ and $$list2$$ respectively.

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  Space complexity : $$O(l_1*x)$$ . hashmap size grows upto $$l_1*x$$ , where $$x$$ refers to average string length.

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Analysis written by: @vinod23