I'm working with an array that contains multiple names such as:
[mike, bob, john, john, rick, bob]I'm working with an array that contains multiple names such as:
[mike, bob, john, john, rick, bob]Create an object with a count attribute by utilizing the Array#reduce method. Afterwards, arrange the array of property names (obtained with the help of Object.keys method) based on the count using the Array#sort method and ultimately retrieve the first element.
var exampleArr = ['mike', 'bob', 'john', 'john', 'rick', 'bob'];
// create an object to store the count
var object = exampleArr.reduce(function(o, v) {
// check if the property is defined
o[v] = o[v] || 0;
// increment the count
o[v]++;
// return the object
return o;
// initialize as an empty object
}, {});
// sort by count in descending order and get the first element
console.log(
Object.keys(object).sort(function(a, b) {
return object[b] - object[a];
})[0]
)
// alternatively, use reduce to find the property with the highest count
console.log(
Object.keys(object).reduce(function(a, b) {
return object[b] > object[a] ? b : a;
})
)To efficiently find duplicates within a list in linear time complexity O(n), one common approach is to iterate through the list and add each item to a set. Before adding an item, check if it already exists in the set. If it does, then that item is a duplicate.
names = ['mike', 'bob', 'john', 'john', 'rick', 'bob'];
seen = new Set();
names.forEach(function(item, index, array) {
if (seen.has(item)) {
console.log(item + " is a duplicate");
} else {
seen.add(item);
}
});
Another method is to sort the list in O(n log(n)) time complexity, saving on extra space usage. While iterating over the sorted array, compare pairs of elements to identify duplicates:
names = ['mike', 'bob', 'john', 'john', 'rick', 'bob'];
names.sort().forEach(function(item, index, array) {
if ((index > 0) && (array[index - 1] == item)) {
console.log(item + " is a duplicate");
}
});
def find_duplicates(lst):
seen = set()
for item in lst:
if item in seen:
print(item)
seen.add(item)
find_duplicates(['mike', 'bob', 'john', 'john', 'rick', 'bob'])To utilize the map function, you can do the following:
function findDuplicates() {
var name, names = ['mike', 'bob', 'john', 'john', 'rick', 'bob'];
var map = new Map();
var max = 1;
var maxRecurringString = "";
for(name of names) {
if(map.get(name) === undefined) {
map.set(name, 1);
} else {
var count = map.get(name);
count = count+1;
map.set(name, count);
if(max < count) {
max = count;
maxRecurringString = name;
}
}
}
console.log("Maximum recurring string is ", maxRecurringString, ". Max number of times :" + max);
}
findDuplicates();This code snippet displays the first string that appears the most number of times. For instance, in the example provided above, both bob and john appear twice. If you wish to print all strings that appear the maximum number of times, you can iterate through the map where the count matches the max count.
If you want to optimize your code for performance, the most efficient approach is to iterate through the array only once.
Avoid using Array#reduce, Array#sort, and Array#forEach as they will loop over the entire array, impacting performance especially with large datasets.
var findHighestRecurring = function(arr) {
arr.sort();
var i = arr.length;
var max = { item:"", count: 0 };
var last = { item:"", count: 0 };
var validate = function() {
if (last.count > max.count) {
max.item = last.item;
max.count = last.count;
}
}
while (i--) {
var curr = arr[i];
if (last.item !== curr) {
validate();
last.item = curr;
last.count = 0;
}
last.count++;
}
validate();
return max;
}
var sample = ["mike","bob","john","bob","john","rick","bob"];
var result = findHighestRecurring(sample);
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