Find next smallest value in array? - arrays

Question

This question has been eating at me for a while. Basically, if you have an input array [4, 5, 2, 6, 7, 1], the next smallest number for 4 and 5 is 2, and for 6 and 7 is 1. I need to identify these smaller numbers. I have an obvious n^2 solution in time, but I feel there is an O(n) solution in time. I need to act on the presence of the next smallest number to the right, and also act in the case there is no smaller number to the right.
I've tried thinking of dynamic programming solutions and data structures (specifically a stack), but I can't seem to keep correctness and O(n) time complexity both in check, one or the other seems to fail for me.
Any help?

Answer 1

You can think of using a stack data structure for this. I have implemented it in Java. The idea is push the index to pop and while the top index value in stack is greater than current index value of the array pop the stack and assign the value at current index to the poped index location.
// Java Implementation of the idea
import java.util.Arrays;
import java.util.Stack;
public class NextSmallest{
public static void main(String[] args) {
int [] A = {4, 5, 2, 6, 7, 1};
int [] ret = nextSmallest(A);
System.out.println(Arrays.toString(ret)); // prints [2, 2, 1, 1, 1, 0]
}
static int [] nextSmallest(int [] A) {
Stack<Integer> stack = new Stack<>();
int n = A.length;
int [] nextSmallestIndex = new int[n];
for(int i = 0; i < n; i++) {
while(!stack.isEmpty() && A[stack.peek()] > A[i]) {
nextSmallestIndex[stack.pop()] = A[i];
}
stack.push(i);
}
return nextSmallestIndex;
}
}

Answer 2

You can use this simple loop over all array elements and get index of numbers you want
var arr = [4, 5, 2, 6, 7, 1];
var object = {};
for (var i = 0; i < arr.length; i++) {
if (i > 0 && i < arr.length - 1 && arr[i] < arr[i - 1] && arr[i] < arr[i + 1]) {
object[i] = arr[i];
} else if (i >= arr.length - 1 && arr[i] < arr[i - 1]) {
object[i] = arr[i];
}
}
console.log(object);
The object for output represents the index of next smallest number and its value. So if you want the smallest value for 4 or 5, because their indexes are before 2, their next smallest number will be object[2] and also for 6 and 7.
Full code
var arr = [4, 5, 2, 7, 6, 1];
var object = {};
for (var i = 0; i < arr.length; i++) {
if (i > 0 && i < arr.length - 1 && arr[i] < arr[i - 1] && arr[i] < arr[i + 1]) {
object[i] = arr[i];
} else if (i >= arr.length - 1 && arr[i] < arr[i - 1]) {
object[i] = arr[i];
}
}
function getResult(inpArr, number, obj) {
var keys = Object.keys(obj);
var index = inpArr.findIndex(function(itm) {
return itm == number;
});
// the number is not in the array
if (index == -1) {
return null;
}
for (var i = 0; i < keys.length; i++) {
if (index == keys[i]) {
return null;
}
if (index < keys[i]) {
return obj[keys[i]];
}
}
}
console.log("Result for 4: ", getResult(arr, 4, object));
console.log("Result for 5: ", getResult(arr, 5, object));
console.log("Result for 6: ", getResult(arr, 6, object));
console.log("Result for 7: ", getResult(arr, 7, object));
console.log("Result for 1: ", getResult(arr, 1, object));
console.log("Result for 2: ", getResult(arr, 2, object));

Answer 3

How do you define your start point? If you know you want to start at element 3 and look from there on, or do you want to find a value, and start at whatever element that value is at and go from there? What happens if a lower value isn't found for the rest of the array?
In PHP just because I'm in the middle of a PHP project, assuming you know what element you want to start with in the array.
<?php
// our array of numbers
$array=array(4, 5, 2, 7, 6, 1);
// what index are we starting at?
$start=2;
// what index are we comparing to?
$counter=0;
// some output
print("First value is ".$array[$start]." at array element ".$start);
// while the initial value is less or equal to the compare value, AND
// the compare position is within the bounds of the array
// increment the counter - our comparison isn't lower
while((($start+$counter)<count($array))&&($array[($start)]<=$array[($start+$counter)])){
$counter++;
}
// did we make it through without exceeding bounds of array or not?
if(($start+$counter)<count($array)){
print("\nNext lower value is ".$array[($start+$counter)]." at array element ".($start+$counter));
}else{
print("\nEnd of array reached without finding lower value...");
}
print("\n");
?>

Answer 4

Find the below solution in O(n) way with optimized code in JavaScript
var arr = [4, 5, 2, 7, 6, 1, 3, 4];
function getNextSmallestValue(index){
var elemvalue = arr[index];
var minValue = elemvalue;
for(var i=index;i<arr.length;i++){
if(minValue> arr[i]){
minValue = arr[i];
}
if((minValue!==elemvalue && elemvalue<arr[i]) || i === arr.length-1){
return minValue;
}
}
}
console.log(getNextSmallestValue(0)); // 2
console.log(getNextSmallestValue(3)); // 1