If the string is sorted in ascending order, the next lexicographically smaller permutation doesn’t exist. Permutes the range [first, last) into the next permutation, where the set of all permutations is ordered lexicographically with respect to operator< or comp.Returns true if such a "next permutation" exists; otherwise transforms the range into the lexicographically first permutation (as if by std::sort(first, last, comp)) and returns false. =, 2 engineers out of 3 engineers and 1 other professional out of 5 professionals can be selected as It changes the given permutation in-place. But this method is tricky because it involves recursion, stack storage, and skipping over duplicate values. (This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.) 4+1 letter can be arranged in 5! If such arrangement is not possible, it must be rearranged as the lowest possible order ie, sorted in an ascending order. The replacement must be in-place, do not allocate extra memory. Permutation with Spaces Easy Accuracy: 45.71% Submissions: 9903 Points: 2 Given a string you need to print all possible strings that can be made by placing spaces (zero or one) in between them. Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. Then T test cases follow. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). Given a word, find lexicographically smaller permutation of it.