There are some problems which do not appear to be a binary search problem at first. For example, ship capacity problem on leetcode. Problem statement is:

A conveyor belt has packages that must be shipped from one port to another within `D` days.

The `i-th` package on the conveyor belt has a weight of `weights[i]`. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within `D` days.

For example:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5Output: 15Explanation: A ship capacity of 15 is the minimum to ship all the packages in 5 days like this: 1st day: 1, 2, 3, 4, 5 2nd day: 6, 7 3rd day: 8 4th day: 9 5th day: 10 Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.

## Ship capacity problem and binary search algorithm

At first glance, it does not appear to be a binary search problem. Where is the input we will search on? That’s where we have to remember that to apply a binary search, we actually do not need an input set, all we need is lower and upper bound.

**Hint 1 **

What is the minimum capacity you would need to ship this cargo? You can choose the lowest possible capacity if you infinite number of days to ship all weights.

**Hint 2**

What if you have only one day to ship all the weights? In that case, what will be the capacity of the ship? Do you need more than capacity ever?

To find these bounds, try to vary the constraint to extremes. In the example of ship capacity, try to put the number of days constraints to an extreme. What if we had an infinite number of days to ship the cargo?

**Can you please explain more?**

Again, what if I had only 1 day to ship the whole cargo? In that case, we need a ship that can take all the weights in 1 day, so ship capacity should be the sum of all the weights.

Now, we know the lower and upper bound of the ship, all we have to adjust the capacity and see if we can ship cargo in D days? Start with mid, see if we can. If yes, then try a smaller capacity. If not, then try greater capacity. All we are doing is find the first capacity between `lower` and `upper` bounds. It seems like the first occurrence problem now.

### Ship capacity problem implementation

public int shipWithinDays(int[] weights, int D) { int upperLimit = 0; int lowerLimit = 0; for(int i = 0; i<weights.length; i++){ upperLimit+= weights[i]; } //Not returning from while loop :) while(lowerLimit < upperLimit){ int shipCapacity = lowerLimit + (upperLimit - lowerLimit)/2; if(isItPossible(D, shipCapacity, weights)){ upperLimit = shipCapacity; } else{ lowerLimit = shipCapacity + 1; } } return lowerLimit; } private boolean isItPossible(int D, int shipCapacity, int[] weights){ int currentLoad = 0; int days = 1; int i = 0; while(i<weights.length){ if(weights[i] > shipCapacity) return false; currentLoad+= weights[i]; if(currentLoad == shipCapacity){ days++; i++; currentLoad = 0; } else if(currentLoad > shipCapacity){ days++; currentLoad = 0; } else{ i++; } } return days <= D; }

The time complexity of the solution is `O(nlogc)` where `n` is number of weights and `c` is capacity range.

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