My solution to a “Daily Coding Problem” that I received in my mail today.

*Given an array of numbers and an index i, return the index of the nearest larger number of the number at index i, where distance is measured in array indices.*

*For example, given [4, 1, 3, 5, 6] and index 0, you should return 3.*

*If two distances to larger numbers are the equal, then return any one of them. If the array at i doesn’t have a nearest larger integer, then return null.*

Here’s my solution in,

```
oneFortyFour(arr: number[], i: number): number | null {
if(arr == null || arr == undefined) {
return null;
}
if(i >= arr.length) {
return null;
}
let distance: number | null = null;
let noAtIdxI = arr[i];
for(let iter=0;iter < arr.length; iter++) {
if(iter == i) continue;
let valAtIter = arr[iter];
if(valAtIter >= noAtIdxI) {
if(distance == null) {
distance = Math.abs(iter - i);
}
else if(Math.abs(iter - i) < distance) {
distance = iter - i;
}
}
}
return distance;
}
```

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