Array Previous Less
Description
Given array of integers, for each position i, search among the previous positions for the last (from the left) position that contains a smaller value. Store this value at position i in the answer. If no such value can be found, store -1 instead.
Example
For items = [3, 5, 2, 4, 5], the output should be
arrayPreviousLess(items) = [-1, 3, -1, 2, 4].
Input/Output
-
[execution time limit] 4 seconds (js)
-
[input] array.integer items
Non-empty array of positive integers.
Guaranteed constraints:
3 ≤ items.length ≤ 15,
1 ≤ items[i] ≤ 200. -
[output] array.integer
- Array containing answer values computed as described above.
[JavaScript (ES6)] Syntax Tips
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// Prints help message to the console
// Returns a string
function helloWorld(name) {
console.log("This prints to the console when you Run Tests");
return "Hello, " + name;
}
Solution
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function arrayPreviousLess(items) {
var extendedItems = [-1].concat(items);
var previousLess = [];
for (var i = 1; i < extendedItems.length; i++) {
var j = i - 1;
while (extendedItems[j] >= extendedItems[i]) {
j--;
}
previousLess.push(extendedItems[j]);
}
return previousLess;
}