var dn = prompt("Enter a number: ");
var bn = new Array();
var i = 0;
var binary = [];
while (dn != 0) {
bn[i] = dn % 2;
dn = Math.floor(dn / 2);
i++;
}
for (var j = i - 1; j >= 0; j--) {
binary.push(bn[j]);
}
console.log(binary.join(""));
var currentGap = 0;
var gaps = [];
var len = binary.length;
for (var k = 0; k < len; k++) {
if (binary[k] == 0) {
currentGap++;
if (binary[k + 1] == 1) {
gaps.push(currentGap);
currentGap = 0;
}
}
}
console.log(Math.max(...gaps));You could add initialize gaps with 0, or add condition when gaps remains empty to print 0 else max of gaps.
var dn = prompt("Enter a number: ");
var bn = new Array();
var i = 0;
var binary = [];
while (dn != 0) {
bn[i] = dn % 2;
dn = Math.floor(dn / 2);
i++;
}
for (var j = i - 1; j >= 0; j--) {
binary.push(bn[j]);
}
console.log(binary.join(""));
var currentGap = 0;
//var gaps = []
var gaps = [0];
var len = binary.length;
for (var k = 0; k < len; k++) {
if (binary[k] == 0) {
currentGap++;
if (binary[k + 1] == 1) {
gaps.push(currentGap);
currentGap = 0;
}
}
}
//if (gaps.length === 0)
// console.log(0)
//else
// console.log(Math.max(...gaps));
console.log(Math.max(...gaps));You can simply print 0 if your gaps array has no length.
var dn = prompt("Enter a number: ");
var bn = new Array();
var i = 0;
var binary = [];
while (dn != 0) {
bn[i] = dn % 2;
dn = Math.floor(dn / 2);
i++;
}
for (var j = i - 1; j >= 0; j--) {
binary.push(bn[j]);
}
console.log(binary.join(""));
var currentGap = 0;
var gaps = [];
var len = binary.length;
for (var k = 0; k < len; k++) {
if (binary[k] == 0) {
currentGap++;
if (binary[k + 1] == 1) {
gaps.push(currentGap);
currentGap = 0;
}
}
}
console.log(gaps.length ? Math.max(...gaps) : 0); // <-- ThisI apologize if I'm misunderstanding your question, but would this satisfy your problem?
const s = prompt("Enter a number:");
const longest = Math.max(...parseInt(s).toString(2).split('1').map(s=>s.length));
console.log(longest);