DEV Community: abhsiksah

Dry running the code to understand the flow

abhsiksah — Sat, 27 Apr 2024 11:29:03 +0000

So I am trying to solve a problem in recursion and backtracking using js as my language. the question is :

` Subsets ( Backtracking Algorithm using Recursion )
// Given an integer array nums of unique elements, return all possible subsets (the power set).
// The solution set must not contain duplicate subsets. Return the solution in any order.

// Input: [1,2,3] ----->>>>> Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
// Input: [0] ----->>>>> Output: [[],[0]]`

My Solution

function subsets(prob) {
  let result = [];
  let temp = [];

  function getPowSet(nums, i) {
    if (i === nums.length) {
      return result.push([...temp]);
    }
    console.log(temp, i, "0");
    temp.push(nums[i]);
    getPowSet(nums, i + 1);
    console.log(temp, i, "1");
    temp.pop();
    getPowSet(nums, i + 1);
    console.log(temp, i, "2");
  }
  getPowSet(prob, 0);
  return result;
}

and I have put certain logs in the code to understand the flow

the logs looks like:-

[ 1 ] 0 0
[ 1, 2 ] 1 0
[ 1, 2, 3 ] 2 0
[ 1, 2 ] 2 1
[ 1, 2 ] 2 2
[ 1 ] 1 1
[ 1, 3 ] 2 0
[ 1 ] 2 1
[ 1 ] 2 2
[ 1 ] 1 2
[] 0 1
[ 2 ] 1 0
[ 2, 3 ] 2 0
[ 2 ] 2 1
[ 2 ] 2 2
[] 1 1
[ 3 ] 2 0
[] 2 1
[] 2 2
[] 1 2
[] 0 2

My concern:

So when the code completes its 1st recursion loop and hits the base case when i become 3 and pushes the temp array to result and returns.. why does the i still logs as 2 in the next log, because there is no i-- going on anywhere inside the base case so it should still log i = 3

here :-

[ 1 ] 0 0
[ 1, 2 ] 1 0
[ 1, 2, 3 ] 2 0
[ 1, 2 ] 2 1 <--------

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abhsiksah — Fri, 16 Jun 2023 05:58:23 +0000

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WHY is this not showing 123???

abhsiksah — Fri, 04 Nov 2022 14:10:31 +0000

var status = 23;

function yzx() {
  var status = 123;

  xyz = () => {
    setTimeout(() => {
      console.log(this.status);
    }, 100);
  };
  xyz();
}
yzx();

Children prop in React!!!

abhsiksah — Sun, 01 Aug 2021 08:25:42 +0000

Refer this below:
https://www.netlify.com/blog/2020/12/17/react-children-the-misunderstood-prop/

Looking for a nice react developer for freelancing in India!!

abhsiksah — Sun, 25 Jul 2021 12:59:56 +0000

Anyone Interested please contact me on 9962258240

Power of JS: reduce

abhsiksah — Sat, 08 May 2021 14:22:16 +0000

// You will be given an array of several arrays that each contain integers and your goal is to write a function that will sum up all the numbers
// in all the arrays. For example, if the input is [[3, 2], [1], [4, 12]] then your program should output 22 because 3 + 2 + 1 + 4 + 12 = 22.
// Solve without and with reduce.

//without reduce

const sumofsubarray = (arr) =>{

let result = 0;
arr.map((elem)=>{

// console.log(elem)
elem.map((eleme)=>{

    result+=eleme;
})


})

return result

}

// console.log(sumofsubarray([[3, 2], [1], [4, 12]]))

//with reduce function

const arr = [[3, 2], [1], [4, 12]]

console.log(arr.reduce((t,e)=>t.concat(e)).reduce((t,e)=>t+e))

So I came across this question.....

abhsiksah — Thu, 06 May 2021 17:50:00 +0000

Given two strings, return true if they are one edit away from each other, else return false.

and this is my solution but I cannot get my answer:

//code:

const isobj = (str) => {
let obj = {};

for (const char of str) {
    !obj[char] ? (obj[char] = 1) : obj[char]++;
}

return obj;

};

const isedit = (str1, str2) => {
let obj1 = isobj(str1);
let obj2 = isobj(str2);

// console.log(obj1,obj2)
let counter = 0;


for (let key in obj2) {
    if (obj1[key] == obj2[key]) 
    {
        continue;
    }
    else {
        if (obj1[key] - obj2[key] == 1) 
        {
            counter++;
        } else
        {
            return false;
        }
    }
}
  console.log(counter);

if (counter == 1) {
    return true;
} else {
    return false;
}

}

console.log(isedit("pal", "pale"));