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Balanced Brackets
Balanced Brackets
+ 0 comments C++14
bool isMatching(char a, char b){ if (a=='{' && b== '}') return 1; else if (a=='(' && b== ')') return 1; else if (a=='[' && b== ']') return 1; else return 0; } string isBalanced(string s) { stack<char> st; for (int i = 0; i < s.length(); i++){ if (s[i] == '(' || s[i] == '[' || s[i] == '{' ){ st.push(s[i]); } else if(st.empty() || !isMatching(st.top(), s[i])){ return "NO"; } else{ st.pop(); } }return st.empty() ? "YES" : "NO"; }
+ 0 comments JS
function isBalanced(s) { let stack = [], brackets = { "}": "{", "]": "[", ")": "(" } for (let str of s.split("")) { if (Object.values(brackets).includes(str)) stack.push(str) else if (stack[stack.length - 1] == brackets[str]) { stack.pop() } else { return "NO" } } return stack.length == 0 ? "YES" : "NO" }
+ 0 comments #python3 def isBalanced(s): stack = [] closeToOpen = { '}': '{', ')': '(', ']': '[' } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return 'NO' else: stack.append(c) return 'YES' if not stack else 'NO'
+ 0 comments JAVA
public static String isBalanced(String s) { // Write your code here Stack<Character> stack = new Stack<>(); for(int i=0; i<s.length(); i++) { if(s.charAt(i) == '(') { stack.push(')'); } else if(s.charAt(i) == '{'){ stack.push('}'); } else if (s.charAt(i) == '[') { stack.push(']'); } else if( stack.isEmpty() || s.charAt(i) != stack.pop() ) { return "NO"; } } return stack.isEmpty() ? "YES" : "NO"; }
+ 0 comments def isBalanced(s): stack=[] for char in s: if char=='{' or char=='(' or char=='[': stack.append(char) if char==']' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='[')): return "NO" if char=='}' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='{')): return "NO" if char==')' and (len(stack)==0 or (len(stack)!=0 and stack[-1]!='(')): return "NO" if char=='}' and len(stack)>0 and stack[-1]=='{': stack.pop(-1) if char==']' and len(stack)>0 and stack[-1]=='[': stack.pop(-1) if char==')' and len(stack)>0 and stack[-1]=='(': stack.pop(-1) if len(stack)==0: return "YES" return "NO"
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