Hello every one

this document is to improve my solving ploblems skill in this document, I will tell how I solve the problems at each problem

like I am a scientist that use sound record while do experiment.

Ha ha if u see this document, please surprise me by texting me the messege at this discord ᲼᲼᲼᲼᲼᲼᲼᲼᲼᲼᲼᲼#7345.

Have a nice day 🙂 Friday 2 December 2022

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Tue 6 December

19:55 - solve count duplicate number in arrays ex13

20:02 - Loop method solve I use Object to count number duplicate number

for (let i = 0; i < arrays.length; i++) {
    dict[arrays[i]] = (dict[arrays[i]] == undefined) ? dict[arrays[i]] = 0 : dict[arrays[i]] = dict[arrays[i]] + 1;
}

then I use Object.keys(`value`) to list all key in Object then run loop again to count value in each key .

key_dict = Object.keys(dict);

answer = 0;

for (let j = 0; j < key_dict.length; j++) {
    answer += dict[key_dict[j]];
}

Thursday 8 December 2022

• 12.00 start doing "count duplicate" by use es6.

• 13.02 finish I use forEach to create object that count the duplicate number of each element and sum all value by method Object.values(object)

• 13.09 start "deleted duplicate" by loop

• 13.26 finish, first I use 2 loop first loop call "main" second loop is "secondary" , in second loop it have condition if (element[main] == element[second] but index of main not equal to secondary) then make that element undefinded

after that we loop again but this loop we append the value that is not undefinded to new answer list.

• 13.39 start "deleted duplicate" by es6
• 13.51 finish "deleted duplicate" by es6

unique = arrays.filter(function(e,i) {
    return arrays.indexOf(e) == i;
});

this is the main point I use filter with indexOf to find check because indexOf will return first index of satisfied element but if that return != current index that mean it not the first element

for instance

[0, 0, 2, 3, 4] 

unique = arrays.filter(function(e,i) {
    return arrays.indexOf(e) == i;
});

first round e, i = 0 arrays.indexOf(e) will return 0 and 0 = i(0) which is true

second round e = 0 , i = 1 arrays.indexOf(e) will return 0 but 0 != i(1) which is false and won't append

I think you might get an idea.