SDE - 1

Use Map data structure with string keys (single characters of the input string).

Inside the for loop check whether the current character has already appeared. If not, just increase the currMaxLen counter by one. Otherwise, if it has appeared earlier and the difference between their 1-based index values loc - prev_loc is less or equal than the current max substring length, the counter must be updated to only contain the length of the new active substring.

For example, for a string "abcb"

after char "c", currMaxLen equals 3 and the map has content "a":1, "b":2, "c":3
for next and last char "b", the if condition results true (2 && 4 - 2 <= 3) and currMaxLen is set to 2 as the active longest substring is now "cb".
iteration ends after that and the answer is three
function lengthOfLongestSubstring(s: string): number {
const map = new Map();

let maxLen = 0;
let currMaxLen = 0;

for (let i=0; i const loc = i + 1;
const prev_loc = map.get(s[i]);

if (prev_loc && loc - prev_loc <= currMaxLen) {
currMaxLen = loc - prev_loc;
} else {
currMaxLen += 1;
}

map.set(s[i], loc);

maxLen = Math.max(maxLen, currMaxLen);
}

return maxLen;
};

class Solution {
public:
vector spiralOrder(vector>& matrix) {
int m = size(matrix), n = size(matrix[0]), sz = m*n;
vector ans(sz);
int r0 = 0, r1 = m-1, c0 = 0, c1 = n-1, r = 0, c = 0;
for (int i = 0, flag = 0; i < sz; flag = (flag+1) % 4) {
if (flag == 0) {
for (c = c0; c <= c1; c++)
ans[i++] = matrix[r][c];
r0++, c--;
} else if (flag == 1) {
for (r = r0; r <= r1; r++)
ans[i++] = matrix[r][c];
c1--, r--;
} else if (flag == 2) {
for (c = c1; c >= c0; c--)
ans[i++] = matrix[r][c];
r1--, c++;
} else if (flag == 3) {
for (r = r1; r >= r0; r--)
ans[i++] = matrix[r][c];
c0++, r++;
}
}
return ans;
}
};

Practice system Designing

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