How can I check if sequence of numbers in an array is increasing by one, or more than one?

Question

I have an array of numbers

let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104];

I need to return 101 because it broke the sequence, it increased by two.

Thanks.

Answer 1

You can start your for loop from the 1 to exclude existing checking.

 let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104]; const findElement = arr => { for(let i = 1; i < arr.length; i++) { if (arr[i] - arr[i-1] > 1) { return arr[i] } } return false; } console.log(findElement(myNumbers)) 

Answer 2

You could use find method and check if prev element is equal to current element - 1.

 let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104]; let result = myNumbers.find((e, i, a) => i != 0 && e - 1 != a[i - 1]); console.log(result) 

Answer 3

Use simple for loop and check value of items.

 let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104]; for(var i=0; i<myNumbers.length; i++) if (i!=0 && myNumbers[i]>0 && myNumbers[i]-1 != myNumbers[i-1]) console.log(myNumbers[i]); 

Answer 4

 let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104]; function incrementsBy(n, sequence) { let lastNumber = sequence[0] for(let n of sequence) { if(n - lastNumber > 1) return n lastNumber = n } return null } console.log(incrementsBy(1, myNumbers)) 

Answer 5

You can use for loop as like below,

 function ShowBreak() { let myNumbers = [95, 96, 97, 98, 99, 101, 102, 103, 104] for (i = 0; i < myNumbers.length - 1; i++) { var currentNo = myNumbers[i]; var nextNo = myNumbers[i + 1]; var differents = nextNo - currentNo; if (differents > 1) { console.log(nextNo); return alert("Sequence break on number " + nextNo + " by increase " + differents); } } return alert('Sequence not breaked.'); } 

 <input type="button" class="btn btn-lg btn-primary btn-block" onclick="ShowBreak()" value="Show" />