Find the biggest interval that has all its members in list in O(n) [duplicate]

Answer 1

Here's a solution similar to Grigor's. Two main differences are that this solution stores the length of the sequential set instead of other indexes and that this eliminates the need for the last hash-set iteration.

1. Iterate over the array

   • Build a hashmap by looking for and updating adjacent set endpoints:

     Key - The array values

     Value - When the key is an endpoint of a sequential set, store the length of that set. Otherwise, keep it truthy so you only consider things once.

   • If the current set size is longest, update the longest set size and longest set start.

Here's a JavaScript implementation for clarity, as well as a fiddle to see it in action:

var array = [1,3,5,7,4,6,10];

//Make a hash of the numbers - O(n) assuming O(1) insertion
var longestSetStart;
var longestSetSize = 0;

var objArray = {};
for(var i = 0; i < array.length; i++){
    var num = array[i];

    if(!objArray[num]){//Only consider numbers once
        objArray[num] = 1;//Initialize to 1 item in the set by default

        //Get the updated start and end of the current set
        var currentSetStart = num;//Starting index of the current set
        var currentSetEnd = num;//Ending index of the current set

        //Get the updated start of the set
        var leftSetSize = objArray[num - 1];
        if(leftSetSize){
            currentSetStart = num - leftSetSize;
        }

        //Get the updated end of the set
        var rightSetSize = objArray[num + 1];
        if(rightSetSize){
            currentSetEnd = num + rightSetSize;
        }

        //Update the endpoints
        var currentSetSize = currentSetEnd - currentSetStart + 1;
        objArray[currentSetStart] = currentSetSize;
        objArray[currentSetEnd] = currentSetSize;

        //Update if longest set
        if(currentSetSize > longestSetSize){
            longestSetSize = currentSetSize;
            longestSetStart = currentSetStart;
        }
    }
}

var longestSetEnd = longestSetStart + longestSetSize - 1;