SDE - Intern

This was famous 3 sum problem.I used hashmap to solve this.This approach uses extra space but is simpler than the two-pointers approach.
Just Run two loops outer loop from start to end and inner loop from i+1 to end. Create a hashmap or set to store the elements in between i+1 to n-1. So if the given sum is x, check if there is a number in the set which is equal to x – arr[i] – arr[j]. If yes print the triplet.

Define a function to calculate the net force at a particular point. This can be done as the definition of the function is given in the question(1/d, where d is the distance).
Take the required inputs
Traverse through the array and apply the binary search algorithm between two adjacent elements.
In the binary search algorithm, we fix a left and a right boundary and then calculate the middle value. We check the net force on the middle point.
If the net force on the middle point is greater than 0 then we move the search space towards the right.
If the net force on the middle point is less than 0 then we move the search space towards right.
If the net force is exactly zero we end the search and move on to find the next equilibrium position between the next pair.

We will declare a map. In that map, we are going to store the value of sum and the current value of an index, if the condition satisfies. Take two variables and set sum to 0 and outputLength to 0. While traversing the array, we will pick each element of an array, and check if arr[i] is equal to 0, if it is found to be equal, we will add -1 to sum and store it to sum, else if we have not found to be 0, we will be adding the positive 1 to sum and store it to sum.

The reason behind that negative 1 and positive 1 is, we are pretending all 0’s to -1 and add them with 1, so we will get the 0 always. But we will check for positive 1 in sum, which indicates that we will have one extra 1 then the count of 0’s.

Suppose, we will take 1, 0, 1 if we are pretending 0 as -1, we will get that 0 with first 2 numbers, and with the third number, we can found that our condition is fulfilled. We got a sub-array of 1’s and 0’s with one extra count of 1 than 0. We get our condition satisfied. That’s why we will be looking for if the sum is equal to 1 in the next step of an algorithm and will be updating the length of outputLength. In the last, if the statement, if we get the new output length, we need to update the previous one with the current outputLength and we will return outputLength

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