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{"@context":"https://schema.org","@type":"Code","url":"https://leetcodejavascript.com/solutions/maximum-sized-array","name":"#3344 - Maximum Sized Array - LeetCode JavaScript Solutions","description":"Given a positive integer s, let A be a 3D array of dimensions n × n × n, where each element A[i][j][k] is defined as:  A[i][j][k] = i * (j O...","programmingLanguage":"JavaScript","codeRepository":"https://github.com/JoshCrozier/leetcode-javascript","codeSampleType":"JavaScript","text":"/**\n * @param {number} s\n * @return {number}\n */\nvar maxSizedArray = function(s) {\n  let left = Math.floor(Math.pow(4 * s, 0.2) / 2);\n  let right = Math.floor(Math.sqrt(2 * Math.sqrt(s))) + 2;\n\n  while (left < right) {\n    const mid = Math.floor((left + right) / 2);\n\n    if (calculateArraySum(mid) > s) {\n      right = mid;\n    } else {\n      left = mid + 1;\n    }\n  }\n\n  return left - 1;\n\n  function getBitOnesCount(n, bitPosition) {\n    const blockSize = 1 << bitPosition;\n    const cycleSize = 1 << (bitPosition + 1);\n    const fullCycles = Math.floor(n / cycleSize);\n    const remainder = n % cycleSize;\n\n    return fullCycles * blockSize + (remainder > blockSize ? remainder - blockSize : 0);\n  }\n\n  function calculateArraySum(n) {\n    let totalSum = 0;\n\n    for (let bitPos = 0; (1 << bitPos) < n; bitPos++) {\n      const onesCount = getBitOnesCount(n, bitPos);\n      const zerosCount = n - onesCount;\n\n      totalSum += (n * n - zerosCount * zerosCount) * (1 << bitPos);\n    }\n\n    return totalSum * (n - 1) * n / 2;\n  }\n};","keywords":"LeetCode 3344, #3344 - Maximum Sized Array, Medium, JavaScript solution","learningResourceType":"Code","isAccessibleForFree":true,"educationalLevel":"intermediate","interactivityType":"mixed"}