ds-algo-study/DS-n-Algos/Dynamic-Programming/tabulation_project/lib/problems.js

Just experimenting with publishing a package

// Write a function, stepper(nums), that takes in an array of non negative numbers.
// Each element of the array represents the maximum number of steps you can take from that position in the array.
// The function should return a boolean indicating if it is possible to travel from the
// first position of the array to the last position.
//
// For Example:
//
// Given [3, 1, 0, 5, 10]
//      - We begin at first position, 3.
//      - Since the element is 3 we can take up to 3 steps from this position.
//      - This means we can step to the 1, 0, or 5
//      - Say we step to 1
//      - Since the element is 1, now the only option is to take 1 step to land on 0
//      - etc...
//
// Try to solve this in two ways, using tabulation and memoization.
//
// Examples:
//
// stepper([3, 1, 0, 5, 10]);           // => true, because we can step through elements 3 -> 5 -> 10
// stepper([3, 4, 1, 0, 10]);           // => true, because we can step through elements 3 -> 4 -> 10
// stepper([2, 3, 1, 1, 0, 4, 7, 8])    // => false, there is no way to step to the end

// Tabulation
function stepper( nums ) {
  const table = new Array( nums.length + 1 ).fill( false );
  table[ 0 ] = true;

  for ( let numIdx = 0; numIdx < nums.length; numIdx++ ) {
    if ( table[ numIdx ] ) {
      for ( let tableIdx = numIdx + 1; tableIdx <= numIdx + nums[ numIdx ]; tableIdx++ ) {
        table[ tableIdx ] = true;
        if ( table[ table.length - 1 ] ) return true;
      }
    }
  }

  return false;
}

// Memoization
function stepper( nums, memo = {} ) {
  if ( nums in memo ) return memo[ nums ]; // Can use nums.length as key to save space complexity
  if ( nums.length === 0 ) return true;
  for ( let nextStep = 1; nextStep <= nums[ 0 ]; nextStep++ ) {
    if ( stepper( nums.slice( nextStep ), memo ) ) {
      memo[ nums ] = true;
      return true;
    }
  }
  memo[ nums ] = false;
  return false;
}

// Write a function, maxNonAdjacentSum(nums), that takes in an array of nonnegative numbers.
// The function should return the maximum sum of elements in the array we can get if we cannot take
// adjacent elements into the sum.
//
// Try to solve this in two ways, using tabulation and memoization.
//
// Examples:
//
// maxNonAdjacentSum([2, 7, 9, 3, 4])   // => 15, because 2 + 9 + 4
// maxNonAdjacentSum([4,2,1,6])         // => 10, because 4 + 6

// Tabulation
function maxNonAdjacentSum( nums ) {
  if ( nums.length === 0 ) return 0;
  let table = new Array( nums.length ).fill( 0 );
  table[ 0 ] = nums[ 0 ];

  for ( let i = 1; i < table.length; i++ ) {
    const useNum = nums[ i ] + ( table[ i - 2 ] || 0 );
    const skipNum = table[ i - 1 ];
    table[ i ] = Math.max( useNum, skipNum );
  }

  return table[ table.length - 1 ];
}

// Memoization
function maxNonAdjacentSum( nums, memo = {} ) {
  if ( nums in memo ) return memo[ nums ];
  if ( nums.length === 0 ) return 0;

  const sums = [];
  nums.forEach( ( num, idx ) => {
    sums.push( num + maxNonAdjacentSum( nums.slice( idx + 2 ), memo ) );
  } );
  memo[ nums ] = Math.max( ...sums );
  return memo[ nums ];
}

// Write a function, minChange(coins, amount), that accepts an array of coin values
// and a target amount as arguments. The method should the minimum number of coins needed
// to make the target amount. A coin value can be used multiple times.
//
// You've seen this problem before with memoization, but now solve it using the Tabulation strategy!
//
// Examples:
//
// minChange([1, 2, 5], 11)         // => 3, because 5 + 5 + 1 = 11
// minChange([1, 4, 5], 8))         // => 2, because 4 + 4 = 8
// minChange([1, 5, 10, 25], 15)    // => 2, because 10 + 5 = 15
// minChange([1, 5, 10, 25], 100)   // => 4, because 25 + 25 + 25 + 25 = 100
function minChange( coins, amount ) {
  let table = new Array( amount + 1 ).fill( Infinity );
  table[ 0 ] = 0;

  coins.forEach( coin => {
    for ( let val = 0; val < table.length; val++ ) {
      for ( let timesUsed = 0; timesUsed * coin <= val; timesUsed++ ) {
        let valNeeded = val - timesUsed * coin;
        let coinsUsed = table[ valNeeded ] + timesUsed;
        if ( coinsUsed < table[ val ] ) table[ val ] = coinsUsed;
      }
    }
  } );
  return table[ table.length - 1 ];
}

module.exports = {
  stepper,
  maxNonAdjacentSum,
  minChange
};