Making a number 1 by adding and dividing
I am having a number n. I have to convert it into 1. The only 2 operations I am allowed to do are dividing it by 2 only if it is divisible by 2 and adding 1. What is the minimum number of steps required to convert the given number into 1. Both adding and dividing operations count as 1 operation.
One method is to follow a greedy strategy. If the number is odd add 1 else divide by 2 until we reach 1. But this is not a correct algorithm.
Another can be to make the number a complete power of 2 by adding the dfference and then keep dividing the number.
But I wonder if any other method might give the best solution
numbers
add a comment |
I am having a number n. I have to convert it into 1. The only 2 operations I am allowed to do are dividing it by 2 only if it is divisible by 2 and adding 1. What is the minimum number of steps required to convert the given number into 1. Both adding and dividing operations count as 1 operation.
One method is to follow a greedy strategy. If the number is odd add 1 else divide by 2 until we reach 1. But this is not a correct algorithm.
Another can be to make the number a complete power of 2 by adding the dfference and then keep dividing the number.
But I wonder if any other method might give the best solution
numbers
add a comment |
I am having a number n. I have to convert it into 1. The only 2 operations I am allowed to do are dividing it by 2 only if it is divisible by 2 and adding 1. What is the minimum number of steps required to convert the given number into 1. Both adding and dividing operations count as 1 operation.
One method is to follow a greedy strategy. If the number is odd add 1 else divide by 2 until we reach 1. But this is not a correct algorithm.
Another can be to make the number a complete power of 2 by adding the dfference and then keep dividing the number.
But I wonder if any other method might give the best solution
numbers
I am having a number n. I have to convert it into 1. The only 2 operations I am allowed to do are dividing it by 2 only if it is divisible by 2 and adding 1. What is the minimum number of steps required to convert the given number into 1. Both adding and dividing operations count as 1 operation.
One method is to follow a greedy strategy. If the number is odd add 1 else divide by 2 until we reach 1. But this is not a correct algorithm.
Another can be to make the number a complete power of 2 by adding the dfference and then keep dividing the number.
But I wonder if any other method might give the best solution
numbers
numbers
asked Nov 15 '18 at 15:30
Nikhil RathoreNikhil Rathore
8319
8319
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53322755%2fmaking-a-number-1-by-adding-and-dividing%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53322755%2fmaking-a-number-1-by-adding-and-dividing%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown